Math 11. Practice Problems from Chapter Fall 01 1 Inverse Functions 1. The graph of a function f is given below. On same graph sketch the inverse function of f; notice that f goes through the points (0, ) and (1, 1). f Solution: The graph of the inverse is the reflection of the given graph over the line =, and it reverse the coordinates of the graph of f, so the graph of f 1 goes through the points (, 0) and ( 1, 1). The graph of f, = and f 1 are given below: f. A function f that is one-to-one on its domain is graphed below.
For our reference, the points plotted on the graph of f are (, 5), (3, ), (0, 3), ( 5, ) (a) Sketch the graph of f 1 on the same graph. (b) Find (i) f 1 ( ) and (ii) f 1 ( 5). (c) Find (i) the domain of f 1 and (ii) the range of f 1. Solution: (a) The graph of f 1 is the reflection of the graph of f over the line =. This is given below. (b) (i) f 1 ( ) = 5 because f( 5) = and (ii) f 1 ( 5) = because f() = 5. (c) (i) The domain of f 1 is [ 5, ) because this is the range of f. (ii) The range of f 1 is (, ] because this is the domain of f. 3. Suppose f is a function and g is its inverse function, suppose also the domain of f is [1, 17] and the range of f is [ 15, 9] with f() = and f(11) = 1. (a) Is f one-to-one? (b) Is g one-to-one? (c) Find the domain and range of g. (d) If possible, find: g(), g(), g(11), g(1). Page
Solution: (a) Yes because f has an inverse function. (b) Yes because g has an inverse function, (its inverse function is f). (c) The domain of g is [ 15, 9] because this is the range of f; and the range of g is [1, 17] because this is the domain of f. (d) For this, use the fact g(b) = a when we know f(a) = b. Therefore, g() = because f() = and g(1) = 11 because f(11) = 1. There is not enough information to find g() and g(11).. Suppose f() = 7 + +. (a) Evaluate: f( 1), f(0), f(1), f(). (b) Given that f has an inverse function, use our answers from (a) to find f 1 (103), f 1 (1), f 1 () and f 1 (). Solution: (a) These are straightforward: f( 1) = + = ; f(0) = ; f(1) = + + = 1 and f() = ()( 7 ) + () + = 103. (b) Recall f 1 (b) = a when f(a) = b. Thus it follows from (a) that f 1 (103) =, f 1 (1) = 1, f 1 () = 0 and f 1 () = 1. 5. Find the inverse function of f() = 17 5 7. Once ou have found f 1 (), verif that f(f 1 ()) = for all. Solution: First write = 17 5 7, and then switch and to obtain = 17 5 7 and solve this for : = 17 5 7 + 7 = 17 5 + 7 17 ( + 7 = 5 = 17 ) 1 5 ( ) 1 + 7 Therefore, f 1 5 () =. 17 We now check [ ( ) ] 1 5 + 7 f(f 1 ()) = 17(f 1 ()) 5 5 7 = 17 7 17 ( ) + 7 = 17 7 = + 7 7 = 17 as desired.. (a) Find the inverse function of f() = 3 + 1, and find its domain. 7 Page 3
(b) Find the range of f. (c) Solve the equation 3 + 1 =, if possible. Note. This is just asking ou to solve 7 f() = for. (d) Solve the equation 3 + 1 7 = 3, if possible. Solution: (a) We first write = 3 + 1, and then switch and to have 7 = 3 + 1 7 Then we will solve the preceding epression for to obtain the inverse function of f. Thus ( 7) = 3 + 1 7 = 3 + 1 3 = 7 + 1 ( 3) = 7 + 1 and so = 7 + 1 3 Thus f 1 () = 7 + 1 3 for 3, and the domain of f 1 = (b) The range of f is the domain of f 1, i.e., the range of f is { : 3 }. { : 3 }. (c) It is possible to solve this, because is in the domain of f 1 (), and f() = is equivalent to = f 1 (), therefore, using (a) = 7() + 1 () 3 = 15 1. (d) This is not possible to solve, because 3 is not in the range of f (see part (b)). 7. Let f() = 5 +. (a) Find f 1 (). (b) Find the domain of f. (c) Find the domain of f 1. (d) Find the range of f. (e) Find the range of f 1. Solution: (a) We first write =, and then switch and to have 5 + = 5 + Page
Then we will solve the preceding epression for to obtain the inverse function of f. Thus Thus f 1 () = (b) The domain of f = (5 + ) = 5 + = = 5 (1 5) = and so = 1 5 1 5 for 1 5. (c) The domain of f 1 = { : 5 }. { : 1 }. 5 (d) The range of f is the domain of f 1 which is (e) The range of f 1 is the domain of f which is { : 1 } 5 { : }. 5. Use the composition propert of inverses to determine if f and g defined below are inverses of each other. ( ) 1 + f() = 1 9 9 and g() = 1 Solution: We check whether (f g)() = for all in the domain of g and (g f)() = for all in the domain of f. First Net [ ( ) ] 1 9 + f(g()) = 1(g()) 9 9 = 1 1 ( ) + = 1 = + = 1 g(f()) = ( f() + 1 = ( 9) 1/9 = ) 1 9 = ( 1 9 + Because (f g)() = for all in the domain of g and (g f)() = for all in the domain of f we know that f and g are inverse functions. 1 ) 1 9 9. Use the composition propert of inverses to determine if f and g defined below are inverses of each other. f() = 10 and g() = 1 10 Page 5
Solution: We first check g(f()) = 1 10(10) = 1 10 Because (g f)(), we know that g and f are not inverses of each other. Page
Eponential Functions 1. Use properties of eponents to solve = (calculators should not be used). Solution: Since > 0, and 1, these epressions are defined, and the eponents must be equal, and so = = + = where the reader can simplif the final fraction when possible.. Consider the function f() = (a) Complete the following table of values for f. 3 1 0 1 3 (b) Sketch a graph of f, and on the same coordinate aes sketch = and the graph of f 1 () (c) Using our answer from (b), sketch g() = +1. Solution: (a) The completed table is as follows. 3 1 0 1 3 1/ 1/ 1/ 1 In (b), recall that the graph of the inverse will be the graph of f reflected over the line =. In (c), the graph of = is shifted 1 units to the left and units down to obtain the graph of g. (b) f f 1 (c) g Page 7
3. Consider the function f() = ( ) 1 3 (a) Complete the following table of values for f. 3 1 0 1 3 (b) Sketch a graph of f, and on the same coordinate aes sketch = and the graph of f 1 () (c) Using our answer from (b), sketch g() = ( 1 3) 3. Solution: (a) The completed table is as follows. 3 1 0 1 3 7 9 3 1 1/3 1/9 1/7 In (b), recall that the graph of the inverse will be the graph of f reflected over the line =. In (c), the graph of = 3 is shifted units to the right and 3 units down to obtain the graph of g. (b) (c) f f 1 g. The graph of an eponential function = b is given on the graph below. Page
(a) Use the graph to estimate b. (b) On the same graph, graph = b 3 5. Solution: (a) b = because the graph goes through the point (1, ). (b) The requested graph results from shifting the original graph 3 units to the right and 5 units down. The graph is given below 5. The graph of an eponential function = b is given on the graph below. Page 9
(a) Use the graph to estimate b. (b) Using the graph from (a), graph = b. (c) Using our graph from (b), graph = b +1. Solution: (a) b = 1 because the graph goes through the point ( 1, ). (b) This graph results in reflecting the part of the original graph for 0 over the -ais. (c) The requested graph results from shifting the original graph 1 units to the left and units down. The graph is given below (b) (c). The graph of an eponential function = b is given on the graph below. Page 10
1 1 1 1 (a) Use the graph to estimate b. (b) On the same graph, graph = b. Solution: (a) b = 1 because the graph goes through the point ( 1, 3). 3 (b) Reflect the graph about the -ais, and then shift it up units to obtain the following graph 1 1 1 1 7. The number of bass in a lake is given b P (t) = 30 1 + e 0.07t Page 11
where t is the number of months that have passed since the lake was stocked with bass. (a) How man bass were in the lake immediatel after it was stocked? (b) How man bass were in the lake 1 ear after it was stocked? Round our answer to the nearest whole number. (c) What will happen to the bass population as t increases without bound? Solution: (a) P (0) = 30 = 0 bass were in the lake immediatel after it was 1+ stocked. 30 (b) P (1) = 9.7 97 bass were in the lake 1 ear after it was 1 + e ( 0.07)(1) stocked. (c) e 0.07t is an eponential function with base b = e 0.07 < 1, this will approach 0 as t increases without bound. Therefore, the bass population will increase, approaching 30 as time increases without bound.. The population of a small cit is currentl 70000 and is growing at 5 percent per ear. Thus the population is given b P (t) = 70000(1.05) t where t is time measured in ears from the present. (a) What will the population of the cit be in one ear? (b) According to this model, what will the population of the cit be in 1 ears from now? Epress answer to the nearest whole number. (c) Suppose Charles has an investment account that is growing a a rate of 5 percent per ear, and he currentl has 70000 dollars in the account. How much mone will be in the account 1 ears from now? Epress answer to the nearest dollar. Solution: (a) P (1) = 70000(1.05) = 73500 which represents an increase of 5 percent over the current populaton. (b) Assuming the current growth rate continues, the population will be P (t) = 70000(1.05) 1 = 13595. (c) This is mathematicall the same question as (b), ecept epressed in dollars, not population, so the account will have 70000(1.05) 1 = 13595 dollars in it after 1 ears. 9. The number e and the natural eponential and logarithm functions. ( (a) The number e is defined b e = lim 1 + 1 n. Use our calculator to complete the n n) following table to get an idea of the approimate value of e: Page 1
Value of n Value of ( 1 + 1 ) n n 10 (1.1) 10 = 100 (1.01) 100 = 1000000 (1.000001) 1000000 = Value of e obtained b using e function on calculator (with = 1): (b) Complete the following table using the e function on our calculator (use decimal places). Then sketch a rough graph of f() = e. 1 0 1 e (c) On the graph given in (b) sketch the following functions: (i) g() = e 3 (ii) h() = e 3 (iii) k() = log e Hints. (i) This is a horizontal translation of the graph of f() = e. (ii) This is a vertical translation of the graph in (i) (iii) log e is usuall denoted b ln and called the natural logarithm. The function k() = ln() is the inverse function of f() = e. Solution: (a) Page 13
Value of n Value of ( 1 + 1 ) n n 10 (1.1) 10 =.5937 100 (1.01) 100 =.70133 1000000 (1.000001) 1000000 =.7117 M calculator value using the e function ke is e.711. The number e is irrational, so it has a non repeating decimal epansion. (b) and (c): e 0.1 1 0.37 0 1 1.7 7.39 10. Eponential functions are ideal for dealing with large numbers (or etremel small numbers), because once we know the base, onl the eponent changes. For eample 10 71.3 and 10 71.57 are numbers larger than man calculators will accept, et properties of eponents make them eas to multipl (or divide). Indeed, 10 71.3 10 71.57 = 10 71.3+71.57 = 10 753.003 Properties of eponents also make it eas to write the above answer in scientific notation 10 753.003 = 10 753+.003 = 10 0.003 10 753 1.031 10 753 Use properties of eponents to answer the following questions. (a) Write 10 1539.99 in scientific notation. Use significant figures in our final answer. Page 1
(b) Find the product 10 1539.99 10 73.71 as a power of 10. (c) Convert our answer in (b) to scientific notation. Use significant digits in our answer. Solution: (a) 10 1539.99 = 10 1539+0.99 = 10 0.99 10 1539 1.9907 10 1539 (b) 10 1539.99 10 73.71 = 10 1539.99+73.71 = 10 303.017 (c) 10 303.017 = 10 303+0.017 = 10 0.017 10 303 1.0399 10 303 Page 15
3 Logarithmic Functions 1. (a) Evaluate log 1. (b) Evaluate log ( 1 3 ) (c) Evaluate log 1 1 Solution: (a) Let = log 1. Then the eponential form is = 1 or = and so =. That is log 1 =. ( ) 1 (b) Let = log. In eponential form, the equation becomes = 1 3 3. Thus ( ) 1 = 1, or = 5 and so = 5. That is log 5 = 5. 3 ( ) 1 (c) From the solution to (a), = 1, and then = 1 and then log 1 1 =.. Find the domain of the logarithmic function f() = log ( 1). Solution: Because the domain of log ( ) is all positive real numbers, the domain of f is all real numbers such that 1 > 0. To solve this, we find the zeros of the quadratic 1 = 0 Therefore, ( 1) = 0 and so = 0 or = 1. 1 > 0 if < 0 or 1 <. Therefore, Using test values, we find domf = { : < 0 or 1 < }. 3. Find the domain of f() = log(9 + ). Solution: The domain of f is the set all real numbers so that 9 + > 0, and therefore 9 > or > 9. Then domf = { : > 9} (where the reader can simplif if necessar). 9. (a) Change the equation log 1 5 = to eponential form. (b) Change the eponential equation 5 = 5 to logarithmic form. Page 1
Solution: (a) ( 1 ) = 5. (b) log 5 5 =. 5. (a) Write the equation log 1 = 3 in eponential form. (b) Write the eponential equation 9 5 = 5909 in logarithmic form. Solution: (a) ( 1 ) 3 =. (b) log 9 5909 = 5.. Consider the functions f() = 5 and g() = log 5 (a) Complete the following table of values for f() = 5. 3 1 0 1 3 5 (b) Complete the following table of values for g() = log 5 b filling in the -values for the given -values. log 5 3 1 0 1 3 What do ou notice about the table of (a) and (b)? (c) Sketch the graphs of f() = 5 and g() = log 5 on the same coordinate aes. (d) Using our answer from (b), sketch h() = log 5 ( + ) 3. Solution: (a) The completed table is as follows. 3 1 0 1 3 5 1/15 1/5 1/5 1 5 5 15 (b) Because g() is the inverse function of f(), it switches the and values from f, so its table just switches the rows of the table in (a). 1/15 1/5 1/5 1 5 5 15 log 5 3 1 0 1 3 In (c), recall that the graph of the inverse g will be the graph of f reflected over the line = which is the dashed line in the graph below. In (d), the graph of g is shifted units to the left and 3 units down to obtain the graph of h. Page 17
(c) f g (d) h 7. Consider the functions f() = ( 1 5) and g() = log 1 5 (a) Complete the following table of values for f() = ( 1. 5) 3 1 0 1 3 ( 1 5) (b) Complete the following table of values for g() = log 1 b filling in the -values for 5 the given -values. log 1 3 1 0 1 3 5 What do ou notice about the table of (a) and (b)? (c) Sketch the graphs of f() = ( 1 5) and g() = log 1 on the same coordinate aes. 5 (d) Using our answer from (b), sketch h() = log 1 ( 1) +. 5 Solution: (a) The completed table is as follows. 3 1 0 1 3 ( 1 ) 15 5 5 1 1/5 1/5 1/15 5 (b) Because g() is the inverse function of f(), it switches the and values from f, so its table just switches the rows of the table in (a). 15 5 5 1 1/5 1/5 1/15 log 1 3 1 0 1 3 5 Page 1
In (c), recall that the graph of the inverse g will be the graph of f reflected over the line = which is the dashed line in the graph below. In (d), the graph of g() is shifted 1 units to the right and units up to obtain the graph of h. (c) f g (d) h. Consider the function f() = log 7 ( + ) (a) Determine the domain of f. Write our answer in interval notation. (b) Solve the equation = log 7 (+) for b converting the equation to eponential form. (c) Use our answer in (b) to complete the following table of values where has been chosen first. 7 5 3 1 (d) With the the help of our table in (c), sketch the graph of f(). Solution: (a) A number is in the domain of f if + > 0, that is when >. Therefore, domf = (, ) (b) Observe that = log 7 ( + ) implies + = log 7 ( + ) and so converting this to eponential form implies (c) Using (b), when + = 7 + = 7 + = 7, = 7 3 = 1 33 = 057 33 =, = 7 = 1 9 = 93 9 Page 19
= 5, = 7 1 = 1 7 = 1 7 =, = 7 0 = 5 = 3, = 7 1 = 1 =, = 7 = 3 = 1, = 7 3 = 337 Using this, the completed table is 057 33 93 9 1 7 5 1 3 337 7 5 3 1 (d) A graph of f() = log 7 ( + ) is as follows. 9. Consider the function f() = log 1 ( + ) + 3 (a) Determine the domain of f. Write our answer in interval notation. (b) Solve the equation = log 1 (+)+ for b converting the equation to eponential 3 form. (c) Use our answer in (b) to complete the following table of values where has been chosen first. 5 3 1 0 1 (d) With the the help of our table in (c), sketch the graph of f(). Page 0
Solution: (a) A number is in the domain of f if + > 0, that is when >. Therefore, domf = (, ) (b) Observe that = log 1 ( + ) + implies = log 1 ( + ) and so converting 3 3 this to eponential form implies (c) Using (b), when + = ( 1 3 ) = = 5, = 3 3 = 1 7 = 53 7 =, = 3 = 1 9 = 17 9 ( ) 1 3 = 3, = 3 1 = 1 3 = 5 3 =, = 3 0 = 1 = 1, = 3 1 = 1 = 0, = 3 = 7 = 1, = 3 3 = 5 Using this, the completed table is 53 7 17 9 5 3 1 1 7 5 5 3 1 0 1 (d) A graph of f() = log 1 ( + ) + is as follows. 3 Page 1
10. The graph of a function f() = log b is given in both graphs below. (a) (b) f f (a) Sketch the graph of g() = log b (b) Sketch the graph of h() = log b (c) Use the given graph to find the base b. Solution: (c) Notice that base is b = because the graph goes through the point (, 1). The requested graphs are as follows. Notice for (a) the graph will be made smmetric about the -ais, and shifted down units. For (b), the graph will be reflected over the -ais and then verticall stretched b a factor of. (a) g (b) h 11. The graph of a function f() = log b is given in both graphs below. Page
(a) f (b) f (a) On the same graph, sketch the graph = and the inverse function of f() = log b. (b) On the same graph, sketch the graph of h() = log 1/b (c) Find the inverse function of f() = log b. Solution: The requested graphs are as follows. Notice for (a), the inverse is obtained b reflecting the graph of f over the line =. For (b), notice log 1/b = log b so the graph of h is obtained b reflecting the graph of f over the -ais. (a) f 1 f (b) h (c) Notice that base is b = 5 because the graph goes through the point (5, 1), and so the inverse function is f 1 () = 5. f 1. The common log of the positive number M is given below. log(m) = 0. (a) Epress M as a power of 10. (b) Epress M in scientific notation. Use significant figures in our answer. Page 3
Solution: (a) log(m) = 0. implies M = 10 0.. (b) M = 10 0. = 10 1+0.35 = 10 0.35 10 1.03 10 1 13. The common log of the positive numbers M and N are given below. log(m) = 30.19 and log(n) = 30.19 (a) Epress M and N as powers of 10. (b) Epress M and N in scientific notation. Use significant figures in our answers. Solution: (a) log(m) = 30.19 implies M = 10 30.19, and log(n) = 30.19 implies N = 10 30.19 (b) M = 10 30.19 = 10 30+0.19 = 10 30 10 0.19 1.55595 10 30, and N = 10 30.19 = 10 31+0.0 = 10 0.0 10 31.77 10 31 1. In each item below, describe how the graphs of the two functions given are related. ( ) 1 (a) g() = and h() = (b) f() = log () and g() =. (c) k() = 3 and g() =. (d) l() = ( ) + 3 and g() =. (e) f() = log () and m() = log 1 (). Solution: (a) h() = so its graph is the graph of g reflected over the -ais. (b) f and g are inverses of each other, so the graph of f is the graph of g reflected over the line =. (c) The graph of k is the graph of g shifted -units to the right and 3 units down. (d) The graph of l is the graph of g stretched verticall b a factor of and then shifted 3 units upward. (e) The graph of m is the graph of f reflected over the -ais. Page
Properties of Logarithms ( ) 5 z 1. Epand log 5 b and write in terms of log b, log b and log b z (assume > 0, > 0, and z > 0). Solution: log b ( 5 z 5 ) = log b ( 5 z 5 ) log b ( ) = 1 log b( 5 z 5 ) log b ( ) = 1 [log b( 5 ) + log b (z 5 )] log b () = 5 log b + 5 log b z log b. ( ) in terms of log 5 z 3 1, log 1,. Use properties of logarithms to full epand log 1 z and number(s); simplif our answer (assume > 0, > 0, and z > 0). log 1 Solution: log 1 ( 5 z 3 ) = log 1 ( ) log 1 ( 5 z 3 ) = log 1 () + log 1 ( ) 1 log 1 ( 5 z 3 ) ( (1 ) ) = log 1 + log 1 () 1 [log 1 ( 5 ) + log 1 (z 3 )] = + log 1 5 log 1 3 log 1 z. 3. Use properties of logarithms to write 3 + log 3 log 7 log z as a single logarithm with coefficient 1. Solution: 3 + log 3 log 7 log z = log ( 3 ) + log () 1 [log ( 3 ) + log (z 7 )] = log () + log ( ) 1 log ( 3 z 7 ) = log ( ) log ( 3 z 7 ) = log ( 3 z 7 ). Page 5
. Write 3 log b ( 5 z 5 ) log b (z ( ) ) + 5 log b as a single logarithm with coefficient 1. Solution: 3 log b ( 5 z 5 ) log b (z ( ) ) + 5 log b = log b ( 5 z 5 ) 3 log b (z ( ) + log b = log b ( 15 z 15 ) log b (z 3 ) + log b ( 5 5 /3) = log b ( 15 z 15 ) log b (z 3 ) + log b ( 5 5 ) log b 3 ( ) ( ) 0 z 15 5 0 z 9 = log b = log 3z 3 b. 3 ) 5 5. Use properties of logs and eponents to evaluate: (a) e 5 ln 3 (b) 1 3 log 1 (c) log ( 10 ) Solution: (a) e 5 ln 3 = e ln 35 = 3 5 = 3 (b) 1 3 log 1 = 1 log 1 (3) = 3 = (c) log ( 10 ) = log ( ) 10 = log ( 0 ) = 0. (Richter Scale) The magnitude of an earthquake of intensit I on the Richter scale is ( ) I M = log where I 0 is the intensit of a zero-level earthquake. (a) Find the magnitude to the nearest 0.1 of an earthquake that has an intensit of I = 511I 0. ( ) I (b) Write the formula M = log in eponential form, and then solve for I. I 0 (c) Use the formula in (b) to find the intensit of an earthquake that measures 5. on the Richter scale. (d) How man times more intense is an earthquake with a magnitude of.5 than an earthquake with a magnitude of 5.1? Epress answer to nearest whole number. (To do this, find the intensit of each earthquake using our formula from (b), and then divide to find the ratio of the intensities). ( ) 511I0 Solution: (a) The magnitude is M = log = log(511) 7.. I 0 I 0 (b) The eponential form is 10 M = I I 0, and so I = 10 M I 0 an earthquake given its magnitude. provides the intensit of Page
(c) From the formula in (b), the intensit is I = 10 5. I 0 39107I 0. (d) M 1 = 10.5 I 0 = 10 (.5 5.1) = 10 3. 51. Therefore, a magnitude.5 earthquake M 10 5.1 I 0 is approimatel 51-times as intense as a magnitude 5.1 earthquake. 7. The range of sound intensities that the human ear can detect is ver large, so a logarithmic scale is used to measure them. The decibel level (db) of a sound is given b db(i) = 10 log I I 0 where I 0 is the intensit of sound that is barel audible to the huma ear. (a) How man times as great is the intensit of a sound that measures 153 decibels when compared to a sound that measures 113 decibels? (b) Find the decibel level of a sound whose intensit is 171 times as intense as a sound that measures 113 decibels. Round our answer to one decimal place. Solution: First observe that the equation db(i) = 10 log I I 0 implies db(i) 10 = log I I 0 I I 0 = 10 db(i)/10 I = I 0 10 db(i)/10 (a) Therefore, the intensit of a sound that measures 113 decibels is I 1 = I 0 10 11.3 ; and the intensit of a sound that measures 153 decibels is I = I 0 10 15.3. The ratio of intensties is I = I 010 15.3 I 1 I 0 10 = 11.3 1015.3 11.3 = 10 = 10000. Therefore, a sound that measures 153 decibels is 10000 times as intense as a sound that measures 113 decibels. (b) From part (a), the intensit of a sound that measures 113 decibels is I 1 = I 0 10 11.3. Therefore, a sound that is 10000 times as intense, has an intensit of I = 171I 0 10 11.3 and its decibel measure will be 10 log(171 10 11.3 ) = 10 log 171 + 113 = 135.3. The ph of a solution with a hdronium-ion concentration [H + ] mole per liter is given b ph = log[h + ]. The higher the concentration of [H + ], the more acidic a solution is considered to be. Therefore, because of the negative sign in the above formula, the lower the ph, the more acidic a solution is considered to be. In particular, a solution with concentration of above 10 7 is considered to be acidic whereas a solution with concentration of below 10 7 is considered to be basic. Page 7
(a) Therefore, a solution with a ph (above/below) is acidic, and a solution with a ph (above/below) is basic. (b) A sample of blood is found to have a hdronium-ion concentration of about 1.90 10 mole per liter. What is its ph? Epress answer to 1 decimal place. (c) A swimming pool s water was tested and the ph level was 7.1. Find the a hdroniumion concentration in mole per liter. Use 3 significant figures in our answer. Solution: (a) A solution with a ph below 7 is considered to be acidic. log(10 7 ) = 7.) A solution with a ph above 7 is considered to be basic. (b) The ph is log(1.90 10 ) = 7.7. (c) The concentration is 10 7.1 = 3.9 10 mole per liter. (Note 9. The percentage of a certain radiation that can penetrate millimeters of lead shielding is given b P () = 100e 1.5 (a) Find the percentage of radiation that can penetrate mm of lead shielding. Use at least 3 significant figures in our answer. (b) How man millimeters of lead shielding are required so that less than 0.013 percent of radiation will penetrate the shielding? Round answer to one decimal place (nearest tenth of a millimeter). Solution: (a) In this case = and so P () = 100e.0 0.775% (b) Solve 0.013 = 100e 1.5 and so e 1.5 = 0.00013 and this implies 1.5 = ln(0.00013) = ln(0.00013) 1.5.0 mm That is, the lead shielding should be approimatel.0 millimeters thick. 10. Which is bigger 0 or 0? Solution: Logarithms are ideall suited for dealing with large numbers, for this question we will take log base 10 of both numbers, and compare which logarithm is larger. log(0 ) = log(0) 101.5 while log( 0 ) = 0 log() 179. Because 101.5 > 179., we know 0 is greater than 0 Page
11. Logarithms are ideall suited for dealing with large numbers, not onl can ou find, for eample, that 75 7 is bigger than 7 75 b comparing logarithms, it is eas to convert a logarithm to scientific notation which gives (perhaps more meaningful) information on the size of a number. The first step is to remember log() = a means = 10 a and properties of eponents then make it eas to convert the number in scientific notation. For eample Therefore, log(75 7 ) = 7 log(75) 7(.3135) = 1.519 75 7 10 1.519 = 10 1+0.519 = 10 1 10 0.519.77 10 1 This is a huge number, it has 15 digits, but scientific notation, at least, provides a scale of its magnitude. Use properties of logarithms and eponents to answer the following questions. (a) Write 10 7.1 in scientific notation. Use significant figures in our final answer. (b) Suppose log(n) = 3.31, write N in scientific notation. (c) Use properties of logarithms, as in the above eample, to write 9 7 in scientific notation. Use significant figures in our final answer. (Use a calculator to check our answer). (d) Use properties of logarithms, as in the above eample, to write 3 193 in scientific notation. Use significant figures in our final answer. (Man calculators cannot process numbers above 10 500, so ou ma not be able to use a calculator to check our answer). (e) Find the product of numbers 9 7 and 3 193. Epress our answer in scientific notation. Use significant figures in our answer. Solution: (a) 10 7.1 = 10 7+0.1 = 10 0.1 10 0.1 1.30 10 0.1 (b) Writing log(n) = 3.31 in eponential form ields N = 10 3.31. Then 10 3.31 = 10 3+0.31 = 10 0.31 10 3.0 10 3 (c) First log(9 7 ) = 7 log(9).70507931. Therefore, 9 7 10.70507931 = 10 0.7050+ 5.075 10 (d) First log(3 193 ) = 193 log(3) 5.5973731530. Therefore, 3 193 10 5.5973731530 = 10 0.597373+5 3. 10 5 (e) You know how to multipl numbers in scientific notation, so ou can use our answer from (c) and (d) and do that! Alternativel, if ou did not have those answers, ou could just work with the logs and then convert the log to scientific notation. log(9 7 3 193 ) = log(9 7 ) + log(3 193 ).70507931 + 5.5973731530 = 93.9731111 Page 9
Therefore, 9 7 3 193 10 93.9731111 = 10 0.9731 10 93 1.971 10 93 1. Logarithms are ideall suited for dealing with large numbers. Use properties of logarithms and eponents to answer the following questions. (a) Use properties of logarithms to write 7 17 in scientific notation. Use significant figures in our final answer. (b) Find the product 7 17 3 1300. Epress our answer in scientific notation. Use significant figures in our answer. (c) Find the quotient 3 1300 7 17. Epress our answer in scientific notation. Use significant figures in our answer. Solution: (a) First log(7 17 ) = 17 log(7) 90.50991. Therefore, 7 17 10 90.50991 = 10 0.5099+90.775 10 90 (b) We will convert the numbers to log base 10, then use properties of logarithms (or eponentials) and then convert number to scientific notation. Therefore, log(7 17 3 1300 ) log(7 17 ) + log(3 1300 ) 90.50991 + 330.053017130555 = 399.911503 7 17 3 1300 10 399.911503 = 10 0.9115 10 399 3.159 10 399 (c) This is similar to (b), ecept for division, ou subtract eponents/logs. For comparison, this solution will use the eponential representation of the numbers, base 10, as converted from logs in (b): 3 1300 7 17 = 10 330.053017130555 10 90.50991 10 330.053017130555 90.50991 = 10 13.0791 = 10 0.0791 10 13.053 10 13 Page 30
5 Eponential and Logarithmic Equations 1. (a) Find the eact solution to (3 ) 7(9 ) = 0. (b) Use a calculator to epress our answer in (a) to five decimal places. Solution: (a) First, (3 ) 7(9 ) = 0 implies (3 ) = 7(3 ) and so (3 ) = 7(3 ). Now divide both sides of this equation b 7(3 ) to obtain 7 = 3 3 and so 7 = 3 Now taking logs of both sides of the latter epression, we have ( ) log = log(3) 7 and so = log(/7) log(3) (This is the eact solution). (b) Using a calculator: = log(/7) log(3) 1.1031.. Solve the equation 7 = 1 3 for. Leave answer in eact form. Solution: Taking the natural log of both sides of the equation and distributing we obtain ( 7)(ln ) = ( 3 )(ln 1) ( ln ) 7 ln = ( 3 ln 1) ln(1) now bring all the s to the left side, and all the numbers to the right side to obtain ( ln + 3 ln 1) = 7 ln ln 1 and then dividing both sides b ln + 3 ln 1 we have = 7 ln ln 1 ln + 3 ln 1 3. (a) Find the eact solution to the equation e 0.e the eact solution. e + e (b) Use a calculator to epress our answer in (a) to si decimal places. = 1. You don t have to simplif 3 Page 31
Solution: (a) Clearing the denominators and then moving like terms to the same sides ields: 3e 0.e = e + e 3e e = 0.e + e (3 1)e = (0. + )e Multipling both sides b e / ields (e )(e ) =. (b) 0.155. e =. = ln ( ). = 1 ( ). ln. Find the eact solution(s) to the equation e + e = 11. Verif that our solutions work. Solution: Viewing the equation as e + e = 11 we multipl both sides of the equation b e to obtain ( (e ) e + ) = 11e e (e ) + = 11e This is an equation of quadratic form, and the substitution u = e turns this into a quadratic equation: u + = 11u u 11u + = 0 (u )(u 3) = 0 This implies u = or u = 3. Then e = implies = ln() and e = 3 implies = ln(3). So the solutions are = ln(), = ln(3). A check of the solutions is as follows: = ln() e ln() + e ln() = e ln() + e ln(1/) = + = + 3 = 11 as desired. = ln(3) e ln(3) + e ln(3) = e ln(3) + e ln(1/3) = 3 + 3 = 3 + = 11 as desired. 5. Solve the equation log 5 ( 3 + 17) = 3. Solution: In eponential form the equation becomes 3 + 17 = 5 3 and so 3 + 17 15 = 0 or 3 + = 0 and this implies ( )( 1) = 0, and so = or = 1. Page 3
. Alwas be careful to check that our solutions work when solving logarithmic equations, because logarithms and their properties are onl defined for positive numbers. See what happens when ou solve the equation: ln() + ln( ) = ln( ) For this, combine the logs on the left side using: ln(m) + ln(n) = ln(mn) when M > 0 and N > 0, and then use the propert ln(e) = ln(f ) implies E = F when ln(e) and ln(f ) are defined. Solution: First, ln() + ln( ) = ln( ) implies ln[( )] = ln( ) and thus ( ) =. Then + = 0 + = 0 ( )( ) = 0 Therefore, = and = are proposed solutions. However, neither of them works because when =, ln( ) = ln(0) which is not defined, and when =, ln( ) = ln( ) which is undefined since < 0. Thus neither of the proposed solutions work, and so there is no solution. 7. Solve the equation log 5 ( + ) + log 5 ( + 7) = log 5 (5 + ). Solution: First, as long as all epressions in the logs are positive, we have log 5 [( + )( + 7)] = log 5 (5 + ) and so ( + )( + 7) = 5 + and that implies and then + 9 + 1 = 5 + and so + 3 = 0 0 = + 3 = ( + )( ) = 0 and so potential solutions are = or =. All the logarithms in the original equation are defined when =, so it is a solution; however when =, log( + ) is not defined since + < 0. Thus = is not a solution. Therefore, the onl solution is =.. The population of a cit is currentl 00000 and is epected to grow at a rate of 5.0 percent per ear for the foreseeable future. Its population is given b P (t) = 00000(1.050 t ) where t is the number of ears from toda. (a) What will the population be in 3 ears? (Epress answer as a whole number) (b) At this rate of growth, how long (in ears) will it take the population to double? How long (in ears) would it take the population to quadruple? Epress answers to 1 decimal place. (c) If this growth rate could continue, how long (in ears) would it take for the population to reach 3,000,000 people? Epress answer to 1 decimal place. Page 33
Solution: (a) The population will be P (3) = 00000(1.050 3 ) 3155 in 3 ears. (b) To determine how long it will take the population to double, we solve (00000) = 00000(1.050) t for t. Thus dividing both sides b 00000 we have = 1.050 t and then log() = t log(1.050) t = log() log(1.050) 1.070 Thus it would take approimatel 1. ears for the population to double. For the population to quadruple, it would have to double twice, so it would take (1.070). ears for the population to quadruple. (c) To determine how long it will take the population to reach 3,000,000, we solve 3000000 = 00000(1.050) t for t. Thus dividing both sides b 00000 we have 3000000/00000 = 1.050 t and then log(3000000/00000) = t log(1.050) t = log(3000000/00000) log(1.050) 55.500 Thus it would take approimatel 55.5 ears for the population to reach 3,000,000 people. Page 3
Applications of Eponentials and Logarithms 1. (Interest Income) Use the properties of logarithms and eponentials, along with the compound interest formula ( A = P 1 + r ) nt n to answer the following questions. (a) Suppose $1000 is invested at an annual interest rate of.0% compounded monthl. How much will it be worth after 5 ears? Epress answer to nearest penn. (b) How long will it take until the investment is worth $110000? Epress our answer in ears rounded to one decimal place. Solution: (a) We use P = 1000 (the amount invested), n = 1 (the number of compounding periods per ear), r =.0/100 = 0.00 (the interest rate in decimal form) and t = 5 (the number of ears the mone is invested). Then ( A = 1000 1 + 0.00 ) (1)(5) 111.37 1 Thus after 5 ears the investment balance will be $111.37 (b) We use P = 1000 (the amount invested), n = 1 (the number of compounding periods per ear), r =.0/100 = 0.00 (the interest rate in decimal form), t (the number of ears the mone is invested is an unknown we wish to determine), and A = 110000 the amount of the investment after t ears. Thus we solve ( 110000 = 1000 1 + 0.00 ) 1t or 1 ( 110000 1000 = 1 + 0.00 ) 1t 1 b taking logs of both sides of the equation to obtain ( log(110000/1000) = 1t log 1 + 0.00 ) t = log(110000/1000) 1 1 log ( ).7131 1 + 0.00 1 Thus it will take approimatel.7 ears for an investment of $1000 to reach a balance of $110000 when invested at.0% compounded monthl.. (Eponential Growth) The population of bacteria in a vat of potato salad at Bob s All Da Buffet is modeled b P (t) = P 0 e kt. At noon there were 1050 bacteria present and at 1:00 pm there were 1750 bacteria present. (a) Find the specific model for P (t) (i.e., use the information given to find P 0 and k, and plug those values into P (t) = P 0 e kt ), and then use it to answer the following questions. (b) How ma bacteria were present in the potato salad at 11:30 am when was placed in the buffet? Epress answer to nearest whole number. (c) How ma bacteria were present in the potato salad at 3:00 pm when the potato salad was removed from the buffet? Page 35
(d) If the potato salad had been allowed to remain in the buffet indefinitel, and the model for the bacteria remained valid, how man hours after noon, when there were 1050 bacteria present would it have taken for the bacteria population to reach 915. Epress answer in hours, rounded to nearest decimal place. (For all of these questions, assume no one ate took an potato salad because of its funn smell and hence the population of bacteria remained in tact and followed the given growth model). Solution: (a) Let t be measured in hours and let t 0 = 0 at 1 noon. Now P 0 = 1050 is the population when t = 0. Thus P (t) = 1050e kt and so we need to find k. Also, we know P (1.0) = 1750 because we are told there were 1750 bacteria present at 1:00 pm, and t = 1.0 at 1:00 pm. Therefore, 1750 = 1050e 1.0k. Then ln(1750/1050) = ln(e 1.0k ) and so 1.0k = ln(1750/1050) k = ln(1750/1050) 1.0 0.5105 Therefore, P (t) = 1050e 0.5105t. (b) At 11:30am, t = 1/ and so there were approimatel 1050e (0.5105)(.5) 13 bacteria present in the vat of potato salad. (c) At 3:00 pm, t = 3.0 and so there were approimatel 1050e (0.5105)(3.0) 1 bacteria in the vat of potato salad. (d) For this, solve P (t) = 915 for t. That is 1050e 0.5105t = 915 and so e 0.5105t = 915 ln ( e 0.5105t) ( ) 915 = ln 1050 1050 ( ) ( ) 915 1 915 0.5105t = ln t = 1050 0.5105 ln.19709 1050 and so it would take approimatel. hours, after noon, for the bacteria population to reach 915. 3. The population of a small cit is currentl 59000 and is growing at percent per ear. Thus the population is given b P (t) = 59000(1.0) t where t is time measured in ears from the present. (a) What will the population of the cit be in one ear? (b) According to this model, what will the population of the cit be in 1 ears from now? Epress answer to the nearest whole number. (c) If this rate of growth continues, how long will it take for the population to triple. Epress answer in ears rounded to two decimal places. Page 3
Solution: (a) P (1) = 59000(1.0) = 010 which represents an increase of percent over the current populaton. (b) Assuming the current growth rate continues, the population will be P (t) = 59000(1.0) 1 = 7. (c) In this case we solve 3(59000) = 59000(1.0) t and so 3 = (1.0) t this implies ln(3) = t ln(1.0) t = ln(3) ln(1.0) 55. The population will triple in approimatel 55. ears.. A population of a town follows an eponential growth model P (t) = Ae kt where t is measured in ears, and P (t) is the population at time t. Suppose the population was 11000 eactl 7 ears ago, and it is 13100 toda. (a) Use this information, with t = 0 for the time eactl 7 ears ago, to find P (t). (b) According to this model, what will the population be 7 ears from now (epress answer to nearest whole number)? Solution: (a) Because P (0) = 11000, we know P (t) = 11000e kt and then P (7) = 13100 implies 13100 = 11000e 7k e 7k = 13100 ( ) 13100 7k = ln k = 1 ( ) 13100 11000 11000 7 ln 11000 Then k 0.095957 and so P (t) 11000e 0.095957t (b) In 7 ears from now, t will be t = 7 + 7 = 1. Then the population will be P (1) 11000e (0.095957)(1) 1501. 5. (Carbon Dating) Carbon-1 has a half-life of 5730 ears, and satisfies the eponentialdeca equation N(t) = N 0 ( 1 ) t/5730. (a) If an ancient scroll is discovered to have 5.0% of its original Carbon-1, how old is the scroll? Round answer to nearest ear. (b) What percentage of a bone s original Carbon-1 would ou epect to find remaining in a bone that is 00 ears old? Round to the nearest tenth of one percent. Page 37
Solution: (a) Solve 0.50N 0 = N 0 (.5) t/5730 and so 0.50 = (.5) t/5730 and so ln(0.50) = t ln(.5) 5730 or t = 5730 ln(0.50) ln(.5) 50 ears Thus the scroll is approimatel 50 ears old. (b) N(t) = N 0 (.5) 00/5730 0.50N 0. This implies that we would epect to find approimatl 5.0% of the original amount of the Carbon-1 remaining in the bone.. An unknown radioactive element decas into non-radioactive substances. In 50 das the radioactivit of a sample decreases b 5 percent, that is percent of the original substance remains. (a) What is the half-life of the element? (b) How long will it take for a sample of 100 mg to deca to 0 mg? Solution: The radioactive deca equation is A(t) = A 0 ( 1 ) t/h where H is the half-life in the units corresponding to t. a) We are given that after 50 das, onl percent of the original sample would remain. Thus we solve This implies 0.A 0 = A 0 ( 1 50 H = ln() ln(.5) ) 50/H and so H = 50 ln(.5) ln(0.) and so the half-life is approimatel 5.53 das. b) The amount of time for a 100 mg sample to deca to a 0 mg sample would be given b ( ) t/5.53 1 0 = 100 or 5.53 ln(0.0) t = 39.7 das ln(.5) 7. According to Newton s Law of Cooling an object placed in a refigerator with a constant temperature of 39 F has its temperature (in degrees Fahrenheit) given b T (t) = 39 + Ce kt where C and k are constants. Suppose a can of soda (the soda is rather warm because the can was in a warm car) had a temperature of 10 F when it was placed in the refrigerator and 0 minutes later, the soda has cooled to F. Page 3
(a) Find the C and k for the temperature equation above. Use t in minutes with t = 0 being the time when the can of soda was placed in the refrigerator. (b) What will be the temperature of the can of soda after 5 minutes? Epress answer to the nearest degree. (c) How long will it take for the can of soda to reach 5 F. Epress answer to the nearest minute Solution: We know T (0) = 10 and so 39 + Ce 0 = 10 and then C = 10 39 = 5. Then T (0) = implies and so k 0.01135. = 39 + 5e 0k e 0k = 9 5 (b) In 5 minutes, the temperature will be approimatel T (5) = 39 + 5e (5)(0.01135) 73 F k = 1 0 ln(9/10) (c) Solve 5 = 39 + 5e 0.01135t and so e 0.01135t = /5 and then 1 t = ln(/5) 19. 0.01135 The soda temperature will reach 5 F approimatel 19 minutes after being placed in the refrigerator.. A cup containing hot tea is placed in a room whose temperature is kept at a constant 7 F. The tea cooled from 05 F to 15 F in 10 minutes. Use Newton s law of cooling to determine how man more minutes, after the temperature reached 15 F, it will take for the tea to cool to 15 F. Epress answer to 3 decimal places. Note: Newton s law of cooling implies T (t) = 7 + Ce kt where t is measured in minutes. Solution: Let t = 0 be the time when the tea is placed in the room. Then T (0) = 05, and so 7 + Ce 0 = 05 and so C = 05 7 = 17. Then T = 7 + 17e kt. Then T (10) = 15 implies 15 = 7 + 17e k(10) and so k = 1 10 ln ( 15 7 17 Therefore, T (t) = 7 + 17e 0.0171353t. ) 0.0171353 To find how long it takes for the temperature to reach 15 F, we solve ( ) 15 = 7 + 17e 0.0171353t 1 15 7 and so t 0.0171353 ln 9.9 17 so it takes an additional 19.9 minutes (after the temperature reached 15 F) for the temperature to reach 15 F. Page 39