Browse · harp

smc

algebra intermediate

Problem

If

P = \frac{s}{( 1 + k ) ^{n}}

then

n

equals:

(A)

f r a c (l o g (s / P), l o g (1 + k))

(B)

l o g (\frac{s}{P ( 1 + k}))

(C)

l o g (\frac{s - P}{1 + k})

(D)

l o g (s / P) + l o g (1 + k)

Solution

P = \frac{s}{( 1 + k ) ^{n}}

(1 + k)^{n} = \frac{s}{P}

Take the

lo g

of each side.

n lo g (1 + k) = lo g (\frac{s}{P})

n = \frac{lo g ( \frac{s}{P} )}{lo g ( 1 + k )} \to \frac{lo g ( \frac{s}{P} )}{lo g ( 1 + k )}

Final answer

A