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algebra senior

Problem

If

lo g_{10} m = b - lo g_{10} n

, then

m =

(A)

b / n

(B)

bn

(C)

1 0^{b} n

(D)

b - 1 0^{n}

(E)

\frac{1 0 ^{b}}{n}

Solution

We have

b = lo g_{10} 1 0^{b}

. Substituting, we find

lo g_{10} m = lo g_{10} 1 0^{b} - lo g_{10} n

. Using

lo g a - lo g b = lo g \frac{a}{b}

, the left side becomes

lo g_{10} \frac{1 0 ^{b}}{n}

. Because

lo g_{10} m = lo g_{10} \frac{1 0 ^{b}}{n}

,

m = \frac{1 0 ^{b}}{n}

.

Final answer

E