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smc

algebra senior

Problem

If

lo g_{8} 3 = p

and

lo g_{3} 5 = q

, then, in terms of

p

and

q

,

lo g_{10} 5

equals

(A)

pq

(B)

\frac{3 p + q}{5}

(C)

\frac{1 + 3 pq}{p + q}

(D)

\frac{3 pq}{1 + 3 pq}

Solution

Notice that

\frac{1}{p} = lo g_{3} 8

, so it would probably be easier to work in base

3

. From change of base,

lo g_{10} 5 = \frac{l o g _{3} 5}{l o g _{3} 10}

. We're given that

lo g_{3} 5 = q

, so now we just need to find

lo g_{3} 10

. We have

\frac{1}{p} = lo g_{3} 8 = 3 lo g_{3} 2

, so

lo g_{3} 2 = \frac{1}{3 p}

. Also,

lo g_{3} 10 = lo g_{3} 5 + lo g 32 = q + \frac{1}{3 p}

. Therefore,

lo g_{10} 5 = \frac{q}{q + \frac{1}{3 p}} = \frac{3 pq}{3 pq + 1}, \frac{3 pq}{1 + 3 pq}

.

Final answer

D