Exercise 128 Page 250

Notice that this is a logarithm with a base of 8. This means if we can rewrite the logarithm's argument as a power of 8, we can determine the result by looking at the exponent.

log_8 8^b = b Since 64 can be rewritten as a power of 8, we can use this identity.

log_8 64 = x

WritePow

Write as a power

log_8 8^2 = x

log_8(8^m)=m

2 = x

RearrangeEqn

Rearrange equation

x = 2

Notice that log_9 x represents the exponent we must raise 9 by to obtain a result of x.

9^(log_9 x)= x Therefore, if we place both sides as exponents on a base of 9, we can eliminate the logarithm.

log_9 x = 1/2

9^(LHS)=9^(RHS)

9^(log_9 x) = 9^(.1 /2.)

9^(log_9(m))=m

x = 9^(.1 /2.)

a^(1/2)=sqrt(a)

x = sqrt(9)

CalcRoot

Calculate root

x = 3

Here, we can use the same rule as we did in Part A to rewrite the logarithm.

log_3 3^4 =x

log_3(3^m)=m

4 =x

RearrangeEqn

Rearrange equation

x=4

Here we can use the same rule as in Part B.

10^(log_(10) 4)=x

10^(log_(10)(m))=m

4=x

RearrangeEqn

Rearrange equation

x=4

In Part C and Part D of this exercise we have the following.

Part C: log_3 3^4 = 4 [0.5em] Part D: 10^(log_(10) 4)= 4 This tells us that a logarithm function and an exponential function undo each other if they have the same base. We can write this generally as follows. log_b b^c = c ⇔ b^(log_b c)= c

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Exercise 128 Page 250

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