Ch 21: Logarithmic Fcts WARM UP You do: 1) Draw the graph of f(x)=2 x. List its domain and range. Is it invertible? Domain: R f(x)=2 x Range: (0, ), or y>0 Yes. It s one-to-one, so it s invertible. Draw its inverse function. Definition: The inverse function of f(x)=2 x is called the logarithm base 2 of x: f 1 x = log 2 (x) The job of log 2 x is to tell you which POWER of 2 equals x: 2? = x. Ex: log 2 1 = 0 because 2 0 = 1, log 2 2 = 1 log 2 1 2 = 1 log 2 4 = 2, etc Since log 2 x undoes 2 x (is its inverse fct): f f 1 x = 2 log 2x = x, and f 1 f x = log 2 2 x = x.
LOG DEFINITION: For any number b that you can use as the base of an exponential function f(x)=b x (i.e, 0<b and b 1) there is a corresponding inverse function called the logarithm base b of x : f 1 x = log b (x) i) If b>1, the graph of log b (x) looks like the one we drew above, so: a. Domain: (0, ), i.e. 0<x. log b x is not defined for negative numbers, since no power of b can be negative. log b x is negative for 0<x<1, positive for x>1, and zero at x=1 b. Range: R c. X-intercept: x=1 d. On the left, as x 0, log b x (vertical asymptote) e. On the right, as x, log b x grows very slowly but it s unbounded ii) How about for 0<b<1? Let s draw the graph of log 1/2 (x). Domain: (0, ) Range: R F(x)=(1/2) x F -1 (x)=log 1/2 (x) On the left, as x 0, log b x (vertical asymptote) On the right, as x, log b x decreases very slowly x-intercept: x=1 The job of log 1/2 x is to tell you the POWER of 1/2 equal to x: (1/2)? = x. Ex: log 1/2 1 = 0 because ( 1 2 )0 = 1, log 1/2 1/2 = 1, log 1/2 1 4 = 2 log 1/2 2 = 1, etc
PROPERTIES OF LOGS: For any A, B >0: 1) log b 1 = 0 2) log b b = 1 (so ln(e)=1) 3) log b AB = log b A + log b B & log b A B = log ba log b B 4) log b A p = p log b (A) 5) b log b x = x & log b b x = x NATURAL LOG function LN(x) Last time we mentioned that the constant e=2.718281828 is very special. In particular, the exp fct f(x)=e x is very special (more about it in Math 124) The inverse fct of e x should be called loge x, but it s used so often that it s called the natural log and it s denoted by ln(x). All scientific calculators have a ln key. note: if you also have a log key, no base specified, it means log base 10 Compute: ln(e)=?, ln(1)=? ln(0)=? ln(e 3 )=?, e ln(5) =? (answers: 1, 0, not defined, 3, 5) 6) Base conversion Formula: log b x = log a x : (in particular: log ln x log a b bx = ) ln b Ex: log 5 7 =ln 7 /ln 5 =1.209061955 (value computed by calculator) log 5 e =ln(e)/ln(5)=1/ln5=0.6213349 (value computed by calculator) EXAMPLES & APPLICATIONS (solutions below): 1) Solve the equations a. 5 3x+2 = 1.2 b. 7 log 3x = 2 5 c. e cos x = 2 2) Rewrite the exponential fct y= 2(5 3x 1 ) in the form y=ae ax for appropriate constants A and a. 3) You need to apply logs to solve for time in problems involving exp modeling. Ex: Pbl 21.9 part c)
Solutions: 1) a. Solve: 5 3x+2 = 1.2 We need to lower the power of 5 so we can solve for x. To do this, you need to use logs. The only choice is which log to start with. One way: Apply the inverse exp function (log base 5) to both sides of the equality: log 5 5 3x+2 = log 5 (1.2) 3x + 2 = log 5 (1.2) (by property 5). Solve this like you would any linear equation. x = 1 3 (log 5 1.2 2) Use the base change formula, then plug in your calculator: x = 1 1.2 (ln 2)= 0.6289057491469 3 ln 5 Another way: use natural log from the beginning: ln(5 3x+2 ) = ln(1.2) (3x+2) ln (5)=ln (1.2) (by property 4 above) 3x+2=ln(1.2)/ln(5) x = 1 1.2 (ln 2)= 0.6289057491469 3 ln 5 Note: Most problems can be solved more than one way. I ll just pick one. b. Solve the equation: 7 log 3x = 2 5 First, since the variable is in the exponent, apply natural log & use properties of logs: Using prop 4 on both sides: Divide by ln 7: ln(7 log 3x ) = ln (2 5 ) log 3 x ln 7 = 5 ln (2) log 3 x = 5 ln 2 / ln 7 = 0.7124143742 To get rid of log base 3, use its inverse function: 3 log 3x = 3 0.7124143742 x = 3 0.7124143742 = 2.187298424
c. Solve the eq: e cos x = 2 ln (e cos x ) = ln (2) cos(x) ln(e)=ln(2). Since ln(e)=1: cos(x)=ln(2) Now you re in trig equation territory: PS: x = cos 1 ln 2 = 0.804950132 SS (for cos): x = cos 1 ln 2 = 0.804950132 Gen Sol: x = ±0.804950132 + 2π k, for all integers k 2) Rewrite the exponential fct y = 2 5 3x 1 in the form y=ae ax for appropriate constants A and a. First, you proceed using the properties of exponents, as if you re trying to put the function in standard exponential form: y = 2 5 3x 1 = 2 5 3x 5 1 = 2 5 53x = 2 5 53 x Now you need to write 5 3 as a power of e. Use the fact that e ln(x) =x, so: 5 3 = e ln (53) = e 3 ln 5 y = 2 5 53 x = 2 5 e3 ln 5 x = 2 5 e(3 ln 5)x So: A=-2/5=-0.4, and a=3ln(5)=4.828313737.
3) First: Exponential Modeling Year zero (t=0) is 1952. We have 2 data points: v(0)=55 thousands and v(10)=75 thousands. We know that v(x) is of the form v(x)=ab x, and we need to first compute the constants A and b. The first data point gives: v(0)=ab 0 =A=55, so A=55. That was easy (no need to solve a system of equations here) The second data point gives: v(10)=55b 10 =75, so b 10 =75/55, so 10 b = 75/55 = 75 55 So: v(x)=55(1.031501485) x 1:10 =1.031501485 WARNING: Remember to not round off the base b too much!!! This gives the value of the house (in thousands of dollars) at x years after 1952. The next question is: When will the house be valued at $200,000? To answer, we need to set v(x)=200 and solve for x: 200=55(1.031501485) x (1.031501485) x =200/55 Since the variable we solve for occurs in the exponent, apply ln to both sides and use the properties of the logs: ln((1.031501485) x )=ln(200/55) x ln(1.031501485)=ln(200/55) x=ln(200/55)/ ln(1.031501485) 41.63 years later (i.e. during 1952+41=1993)