Pre-Calculus Notes from Week 6

Transcription

1 1-105 Pre-Calculus Notes from Week 6 Logarithmic Functions: Let a > 0, a 1 be a given base (as in, base of an exponential function), and let x be any positive number. By our properties of exponential functions, we know that the function f(t) = a t has range (0, ), all positive numbers, so x is in the range of a t. That means there is a y with a y = x. And, because f(t) = a t is either always increasing or always decreasing, we know that if there s a z with a z = x, then a y = a z, and hence y = z. What this means is that there s only one number y with a y = x. Definition: Let a > 0, a 1, and x > 0 be given. Let y be the unique real number such that a y = x. (We just showed y exists in the previous paragraph.) We call y the logarithm of x with base a, and we write y = log a (x). Example: Find log (8). Solution: We can write 8 = 3, so 3 is the number we raise to in order to get 8, so log (8) = log ( 3 ) = 3. Example: Find log 3 (1/9). so Solution: 1 9 = 1 3 = 3, ( ) 1 log 3 = log 9 3 (3 ) =. Example: Find log 7 (7 15 ). Solution: 7 15 is already written as a power of 7, so log 7 (7 15 ) = 15. From how we defined f(x) = log a (x) (where a is any base), we can see that the domain is (0, ) (we only can put positive numbers in) and the range is (, ) (we can get any number out), which is the opposite of what g(x) = a x has. This isn t a coincidence! Exponential Functions and Logarithmic Functions are Inverse to Each Other! For a > 0, a 1, let f(x) = log a (x) and let g(x) = a x. Then, for any number x, (f g)(x) = f(g(x)) = f(a x ) = log a (a x ) = x,

2 and for any x > 0, (g f)(x) = g(f(x)) = g(log a (x)) = a log a (x) = x. (Remember, we defined log a (x) to be the number you raise a to in order to get x out.) So f and g are inverse functions by our very definition: what this means for us is that exponential functions and logarithms can be used to cancel out each other, which is how we ll use them when solving problems. Notation: Just as we wrote sin x instead of sin(x) and cos x instead of cos(x), we ll often write log a x instead of log a (x), when the meaning won t be lost. Important Properties of Logarithmic Functions: Assume a > 0, a log a (1) = 0. This is because a 0 = 1 for all a.. log a (a) = 1, since a = a For A, B, both positive (and here A a: lower-case letter and capital letters are different variables in most math textbooks!), we have log a (A B) = log a A + log a B and ( ) A log a = log B a A log a B. To see why this is true, let s let r = log a A and s = log a B, so A = a r and B = a s. Then A B = a r a s = a r+s and A B = ar a s = ar s, and by the definition of the logarithm, that means that log a (A B) = r+s and log a ( A B ) = r s. Substituting log a A in for r and log b B in for s gives us the formulas. 4. For A > 0, k any number, log a (A k ) = k log a (A). To see why this is true, again let r = log a A, so A = a r. Then A k = (a r ) k = a r k, so log a (A k ) = r k, and the result follows from substituting log a A in for r. 5. If a > 1, the graph looks like

3 If 0 < a < 1, the graph looks like Graph of y = log a (x), a > 1. Graph of y = log a (x), 0 < a < 1. (You may notice a similarity to the graphs of a x for a > 1 and 0 < a < 1 here. This is because the functions log a (x) and a x are inverse functions. Though I didn t discuss it in class, inverse functions have the property that each looks like the graph of the other, but flipped over the line y = x. You can see more of this in your textbook in Section of Appendix B, but I won t be testing on it.) 6. From the graphs, we get that if a > 1, then log a (x) is strictly increasing, and log a (x) gets very, very negative as x gets close to 0, and gets positive (slowly) as x gets larger. If 0 < a < 1, log a (x) is decreasing, and becomes very, very positive as x gets close to 0, and becomes negative (slowly) as x gets larger. Since log a (1) = 0, the x-intercept of log a (x) is always 1.

4 Before we go further, we need two definitions: Definition: We define the common logarithm to be the function f(x) = log 10 (x), the logarithm with base 10, and we write log(x) instead of log 10 (x). Definition: We define the natural logarithm to be the function f(x) = log e (x), the logarithm with base e, where e is the natural exponential base. We write ln(x) instead of log e (x). The ln comes from logarithme naturel, which is, of course, French for natural logarithm. Just as we have Euler (pronounced oiler, not yooler ) to blame for the choice of the letter e for the natural exponential base, so too do we have him to blame for this notation. Euler was Swiss, so it s very likely he spoke French. Note: For various reasons, older textbooks (many textbooks written pre-1970), and some newer textbooks, use log(x) to denote the natural logarithm and not the common logarithm. While this isn t a common occurrence, when you see log(x) written in books, make sure you know which one the author is referring to! To recap: for x > 0, and for any x, e ln(x) = x and 10 log x = x, ln(e x ) = x and log(10 x ) = x. Example: Solve 9 x +x = 1 using logarithms. Solution: Since we want to use logarithms, we look to see if there s a good base to pick: since we have 9 x +x, we guess that 9 might be a good choice. Since we re solving for x, we want to bring it out of the exponent, so we take the logarithm (with base 9) of both sides: 9 x +x = 1, so log 9 (9 x +x ) = log 9 (1), and by properties of logarithms, the left-hand side becomes x + x and the right-hand side becomes 0, leaving us with x + x = 0. Factoring an x out gives us x(x + 1) = 0, so x = 0 or x = 1. Checking our answers, we see that = 9 0 = 1 and 9 ( 1) +( 1) = = 9 0 = 1,

5 so our solutions are x = 0 and x = 1. Notice that this is the same answer we get as if we had used our previous method of substituting 9 0 in for 1, so 9 x +x = 9 0 means x + x = 0 by the equality of exponents. Logarithms, it will turn out, let us not only solve the exponential functions we ve already seen, but many that are more complicated. Example: Solve ln x + ln(x + 1) = 0. Solution: By our logarithm rules, since ln(x) and ln(x + 1) have the same base, ln x + ln(x + 1) = ln(x(x + 1)) = ln(x + x), so ln(x + x) = 0. Since we want to get rid of the ln and just deal with x, and since e is the base of the logarithms we re dealing with, we raise e to both sides: e ln(x +x) = e 0. Since e x and ln(x) are inverse functions, e ln(x +x) = x + x, and we know e 0 = 1, so we re left with x + x = 1, or x + x 1 = 0. Since this is a quadratic, we can use the quadratic formula to solve: x = (1) ± (1) 4(1)( 1) (1) = 1 ± 5, so we get x = 1 5 and x = But our original equation is ln(x) + ln(x + 1) = 0, so we must have x > 0 and x + 1 > 0 (the domain of ln is positive numbers), so x > 0 and x > 1 must both be true, and we see that x > 0 is the most restrictive of the two. But 1 5 < 0 (it s the opposite of the golden ratio), so it can t be a solution to this equation. But since 5 > 4 = > 1, therefore > = 0, 1+ 5 > 0 and therefore it is a solution. So x = Remember, we can only put positive numbers into logs! This makes solving logarithmic equations slightly trickier than solving regular equations, since now we have to make sure that everything inside the log stays positive! Example: Solve log 9 x = 3. Solution: Since we want to get rid of the log 9, we raise 9 to both sides: 9 log 9 x = x and 9 3/ = ( 9) 3 = 3 3 = 7, so x = 7,

6 and we re done. Change of Base Suppose we can calculate log a (x) very easily, but what we actually want to find is log b (x), where a and b are different bases (so both are positive and neither is equal to 1). We can do this, but it requires a little finesse: First, we note that x = b log b (x) since b x and log b x are inverse functions. Then, we note that b = a log a (b) for essentially the same reason. So But this means that x = b log b (x) = (a log a (b) ) log b (x) = a log a (b) log b (x). log a (x) = log a (b) log b (x). Since log a (b) 0 (as b 1), this means we can solve for log b (x), which gives us the change of base formula: log b (x) = log a(x) log a (b). Why would we want to know how to do this? Well, one reason is that computers can calculate binary logarithms (log (x)) very quickly as a result of how they re built, and this formula lets us calculate log a (x) for any a by simply computing log (x)/ log (a). But a more important result is that log a (x) = ln(x) ln(a), which means that every single logarithmic function is a multiple of ln(x)! So we only need to understand the properties of ln(x) and we ll get the properties of log a (x) as a result. Similarly, since a = e ln(a) for every a, we have a x = ( e ln(a)) x = e ln(a) x, so every single exponential function can be formed from e x, and it s enough to understand the properties of e x to figure out the properties of a x. Example: Find log 4 ( 178 ). Solution: Looking at 178, we see it s written as to some power, which means that taking the logarithm with base of it is easy. What that suggests is we should use our change of base formula: log 4 ( 178 ) = log ( 178 ) log (4).

7 So log ( 178 ) = 178 by the fact that x and log (x) are inverse functions, and since 4 =, we have log (4) =, so log 4 ( 178 ) = 178 = 89. Now, let s look at some more general equations, and see how to solve them: Example: Solve 5 x+1 = 6 x. Solution: Since neither 5 nor 6 can be written as a simple power of each other, it doesn t look like there s a good choice of base for logarithms. But we do want to apply logarithms, since we know they undo exponential functions, so we ll use the common log: log(5 x+1 ) = log(6 x ). We might be tempted to move everything to one side of the equation and simplify using our rules for differences of logs, but what we ll find is that, if we do that, it doesn t make the problem any easier. What we ll do instead is use Property 4 of logarithms: log a (A k ) = k log a (A). Applying this idea to both sides, we get (x + 1) log(5) = (x ) log(6). Distributing the log(5) and log(6) terms, we get ( log(5))x + log(5) = (log(6))x ( log(6)), and since log(5) and log(6) are just fixed numbers (albeit complicated ones), we see that this new equation is a linear equation in x, so we can solve it by moving the x terms to the left, the constant terms to the right, and dividing by x s coefficient: ( log(5))x + log(5) = (log(6))x ( log(6)), ( log(5))x (log(6))x = log(5) ( log(6)), ( log(5) log(6) ) x = log(5) log(6), x = log(5) log(6) log(5) log(6). Now, the right-hand side of our value of x looks complicated and unpleasant, but what s important is that there are no x s over there: the right-hand side is a specific number, though it s not pleasant to compute by hand. However, just to make it look a little nicer, let s simplify the numerator and denominator some: log(5) log(6) = (log 5 + log 6) = (log 5 + log 6 ) = (log(5 6 )) = log(180), and ( ) 5 log(5) log(6) = log(5 ) log(6) = log = log 6 ( ) 5, 6

8 so our final answer is x = log(180) log( 5). 6 Example: Solve 5 x 5 x = 3. Solution: This is by far the trickiest problem we ve looked at, because the left-hand side can t clearly be simplified into a single constant to some power. In fact, taking logs here won t help: we need to simplify some other way. The first thing we ll do is multiply both sides by, to get 5 x 5 x = 6. While that doesn t seem to have helped, we can now rewrite the left-hand side: 5 x 5 x = 5 x 1 5 x, and we ll multiply through by 5 x (which is always positive) to get rid of the fraction, and simplify some: (5 x 5 x )5 x = 6 5 x (5 x )(5 x ) (5 x )(5 x ) = 6 5 x (5 x ) 5 x+x 6 5 x = 0 (5 x ) x = 0 (5 x ) 6 5 x 1 = 0. While we could have written (5 x ) as 5 x, writing it like this gives us the missing clue: let u = 5 x. Then our equation is u 6u 1 = 0, a quadratic! We can use the quadratic formula (just like when we solved higher-degree polynomial equations) to get u = ( 6) ± ( 6) 4(1)( 1) (1) = 6 ± 40 = 6 ± 4 10 = 3 ± 10. But u = 5 x > 0, and 3 10 < 0, so the only solution is u = (Remember, we have to pay more attention to the signs of numbers when solving exponential and logarithmic equations!) Therefore 5 x = ,

9 and we can solve this by taking the natural log of both sides: ln(5 x ) = ln(3 + 10), x ln(5) = ln(3 + 10), x = ln(3 + 10). ln(5) Right now, you might be asking yourself, But why didn t we just take log 5 of both sides, since we had 5 x on the left?. Well, the answer is: no reason at all: log 5 (5 x ) = log 5 (3 + 10), so x = log 5 (3 + 10). While this answer isn t identical to ours in how it s written, the two are equal because of the change of base formula: log 5 (3 + 10) = ln(3 + 10). ln(5) Example: Solve log( 3 x) = log x for x. Solution: The first thing we notice is that 3 x = x 1/3. This means that log( 3 x) = log(x 1/3 ) = (1/3) log(x), giving us 1 3 log(x) = log(x). Squaring both sides to get rid of the radical gives us 1 9 (log(x)) = log(x), or 1 9 (log(x)) log(x) = 0. Factoring out log(x) from the left-hand side gives us ( ) 1 log(x) 9 log(x) 1 = 0, so log(x) = 0 or (1/9) log(x) 1 = 0. If log(x) = 0, then taking 10 to both sides gives us x = 10 log(x) = 10 0 = 1, so x = 1. Now, we check to make sure it works in our original equation: log( 3 1) = log(1) = 0, and log(1) = 0 = 0, so both sides are defined and equal: x = 1 is a solution.

10 If (1/9) log(x) 1 = 0, then log(x) = 9 and therefore x = 10 9 : log( ) = log(10 9/3 ) = log(10 3 ) = 3 and log(109 ) = 9 = 3, so x = 10 9 is also a solution.