Exercise 70 Page 340

We want to solve an equation involving more than one logarithm. To do so, we will use the Change of Base Formula to evaluate the given equation. log_b m = log_c m/log_c b In the above formula, m, b, and c are positive numbers, with b≠ 1 and c≠ 1. We will apply this formula with c = 2 to simplify the expressions. Next, we will use the fact that if two equivalent logarithmic expressions have the same base, then the arguments must be equal. log_b x=log_b y ⇔ x= y Let's apply the above property to our equation. log_2 x^3 = log_2 (4x)^2 ⇔ x^3 = (4x)^2 Let's now solve the equation x^3 = (4x)^2.

x^3 = ( 4x )^2

PowProd

(a * b)^m=a^m* b^m

x^3 = 16 x^2

SubEqn

LHS-16x^2=RHS-16x^2

x^3 - 16 x^2 = 0

FactorOut

Factor out x^2

x^2(x-16) = 0

ZeroProdProp

Use the Zero Product Property

lcx^2=0 & (I) x-16=0 & (II)

SqrtEqn

(I): sqrt(LHS)=sqrt(RHS)

lx=0 x-16=0

AddEqn

(II): LHS+16=RHS+16

lx=0 x=16

To check our answer, we will substitute both 0 and 16 for x in the given equation one at a time.

log_4 x = log_8 4x

Substitute

x= 0

log_4 0 ? = log_8 4( 0)

ZeroPropMult

Zero Property of Multiplication

log_4 0 ? = log_8 0 *

Note that you can never get zero by raising a number different from zero to any power, so both log_4 0, and log_8 0 are undefined. Therefore, 0 is an extraneous solution. Let's now check for x = 16.

log_4 x = log_8 4x

Substitute

x= 16

log_4 16 ? = log_8 4( 16)

Multiply

Multiply

log_4 16 ? = log_8 64

Calculate logarithm

2 = 2 ✓

Since substituting 16 for x in the given equation produces a true statement, x=16 is the solution to our equation.