If a=log_8 225 and b=log_2 15, then a, in terms of b, is:

1962 AMC 12 Problem 17

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17.

a=\log_8 225

and

b=\log_2 15,

then

a,

in terms of

b,

is:

\dfrac b2

\dfrac{2b}{3}

b

\dfrac{3b}{2}

2b

Write

225=15^2

and

8=2^3

Use the change-of-base formula with base

2

Changing to base

2

gives

\begin{aligned} a&=\frac{\log_2(15^2)}{\log_2(2^3)}\\ &=\frac{2\log_2 15}{3} =\frac{2b}{3}. \end{aligned}

Thus, the correct answer is B.