Compare these two functions and state how they are related to exponential functions: y = log(x) and y = ln(x)

(a) if log(x) = y, then 10 y = x (b) if ln(x) = y, then e y = x Note x > 0 (i.e. ) in both graphs and When x = 1, y = 0 So ln(1) = log(1) = 0	Graph of both Functions
Inverse functions - y=10 x and y=log(x) Each is the inverse function of the other.	Inverse Functions - ex and ln(x) Each is the inverse function of the other

Example 2

Solve for t in the exponential functions: (a) 12=3(1.05^{) t} and (b) ½ =3(0.85) ^t

(a) 12=3(1.05) t ,  So	(b) ½ =3(0.85) t So
(a) Graphic Solution to 12=3(1.05) t	(b) Graphic Solution to ½ =3(0.85) t

Example 3

When will an investment of $200,000 reach $1,000,000 if it is invested in an fund account that is compounded daily
at an annual interest rate of 6%?

Algebraic Solution So balance, B of the account after t years is: So to find when  Use logs to solve for t: , take logs of both sides: ,  years

Graphic Solution

Example 4

How long would it take the number of microorganisms to double in a jar if the number is growing at a rate of 14.5% each hour?

Algebraic Solution Since exponential growth:  Number doubles when , So , divide by N0 , take logs:  Therefore

Numeric Solution Let initial number be 25, then , So , divide by 25 , taking logs:  Therefore  Check

Example 5

When will the population size of two populations be equal if population A grows from 25,000 at a rate of 0.8% each year
and population B grows from 20,000 at a rate of 0.9% each year?

Algebraic Solution Let PA = PB:  Group like terms:  Note: , so : , take logs

Graphic Solution

Example 6

Solve for x in the following; (a) 2 = log(x-3) and (b) 6 = ln(104 + x)

(a) 2 = log(x-3) Since when y = log(x), , then , So  Check: log(103 - 3) = log(100) = 2 So x = 103

(b) 6 = ln(104 + x) Since when y = ln(x), , then , So  Check: ln(104+299.4288) = 6 So x = 299.4288

Example 7

Solve for x in the following: (a) 1 = log(x-2) + log(x+2) and (b) 2 =ln(10x) - ln(x+2)

(a) 1 = log(x-2) + log(x+2) Properties:  So 1 = log[(x-2)(x+2)]=log(x2 - 4) 1 = log(x2 - 4) Since when y = log(x), , then 101 =x2 - 4, So x2 = 10 + 4 = 14,  Check: 1 So

(b) 2 =ln(10x) - ln(x+2) Properties:  So  Since when y = ln(x), , then ,  Group like terms:  , So x = 5.6601 Check: ln(56.6) - ln(5.66 + 2) = 2

Example 8

Evaluate the following algebraically:

(a) Find x if log(x)=ln(x) (b) log(10,000) (c) log(0.001)

(d) lne^3x+2 (e) log(10) - log(0.01) (f) 10^log100

(a) Find x if log(x)=ln(x) The only time both functions are equal is when x = 1 So log(1) = ln(1) = 0 See Graph above	(b) log(10,000) Note 10,000 = 104 Since log 10x = x Then log 104 = 4	(c) log(0.001) Note  Since log 10x = x Then log 10 -3 = -3
(d) lne3x+2 Since ln ex = x Then lne3x+2 = 3x + 2	(e) log(10) - log(0.01) Note: log 10 = 1, log 0.01=-2 Also:  So log(10)-log(0.01)=1-(-2)=3	(f) 10log100 Note log 100 = log 102 = 2 So 10log100 = 102 = 100 Therefore: 10log100 = 100  Also 10log(x) = x