What is the answer to solvefor n,a=p(1+r)^n ?

The answer to solvefor n,a=p(1+r)^n is n=(ln(a/p))/(ln(1+r))

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solvefor n,a=p(1+r)^n

solve for

n, a = p (1 + r)^{n}

n = \frac{ln ( \frac{a}{p} )}{ln ( 1 + r )}

Solution steps

a = p (1 + r)^{n}

Switch sides

p (1 + r)^{n} = a

Divide both sides by

p

(1 + r)^{n} = \frac{a}{p}

Apply exponent rules

n ln (1 + r) = ln (\frac{a}{p})

Solve

n ln (1 + r) = ln (\frac{a}{p}) : n = \frac{ln ( \frac{a}{p} )}{ln ( 1 + r )}

n = \frac{ln ( \frac{a}{p} )}{ln ( 1 + r )}

2^{y} = 8

2^{y} = 3

2^{y} = 5

2^{x + 2} - 2^{x - 1} - 2^{x - 2} = 26

\frac{2 ^{x}}{512} = (2^{x})^{4}

What is the answer to solvefor n,a=p(1+r)^n ?
The answer to solvefor n,a=p(1+r)^n is n=(ln(a/p))/(ln(1+r))