CK-12-Pre-Calculus Concepts

3.6. Exponential Equations http://www.ck12.org

30 =(^1.^06 )

t− 1

0. 06

1. 8 = ( 1. 06 )t− 1

2. 8 = 1. 06 t

ln 2. 8 =ln( 1. 06 t) =t·ln( 1. 06 )

t=ln 1ln 2.. 068 ≈ 17. 67 years

Example B
Solve the following equation forx: 16x= 25
Solution: First take the log of both sides. Then, use log properties and your calculator to help.

16 x= 25

log 16x=log 25

xlog 16=log 25

x=log 25log 16

x= 1. 61

Example C
Solve the following equation for all possible values ofx:(log 2 x)^2 −log 2 x^7 =−12.
Solution: In calculus it is common to use a small substitution to simplify the problem and then substitute back
later. In this case letu=log 2 xafter the 7 has been brought down and the 12 brought over.

(log 2 x)^2 −7 log 2 x+ 12 = 0

u^2 − 7 u+ 12 = 0

(u− 3 )(u− 4 ) = 0

u= 3 , 4

Now substitute back and solve forxin each case.

log 2 x= 3 ↔ 23 =x= 8

log 2 x= 4 ↔ 24 =x= 16

Concept Problem Revisited
In the equation, logs can be used to reduce the equation to 2x=6.

1. 798982 x= 1. 798986
   Take the log of both sides and use the property of exponentiation of logs to bring the exponent out front.