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$_2$ 8...

Evaluate log $_2$ 8 – log $_3$ $\frac{1}{9}$ ...

Evaluate log $_2$ 8 – log $_3$ $\frac{1}{9}$

A.

-1 1 $\frac{1}{2}$

B.

-1

C.

1

D.

5

View answer & explanation

Correct answer is D

log $_2$ 8 – log $_3$ $\frac{1}{9}$

= log $_2$ 2 $^3$ – log $_3$ 9 $^{-1}$

= log $_2$ 2 $^3$ – log $_3$ 3 $^{-2}$

Based on law of logarithm

= 3 log $_2$ 2 – (-2 log $_3$ 3)

But log $_2$ 2 = 1,

log $_3$ 3 = 1

So, = 3 + 2

= 5

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