What is the answer to solvefor n,F=P(1+i)^n ?

The answer to solvefor n,F=P(1+i)^n is n=(ln(F/P))/(ln(1+i))

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solvefor n,F=P(1+i)^n

solve for

n, F = P (1 + i)^{n}

n = \frac{ln ( \frac{F}{P} )}{ln ( 1 + i )}

Solution steps

F = P (1 + i)^{n}

Switch sides

P (1 + i)^{n} = F

Divide both sides by

P

(1 + i)^{n} = \frac{F}{P}

Apply exponent rules

n ln (1 + i) = ln (\frac{F}{P})

Solve

n ln (1 + i) = ln (\frac{F}{P}) : n = \frac{ln ( \frac{F}{P} )}{ln ( 1 + i )}

n = \frac{ln ( \frac{F}{P} )}{ln ( 1 + i )}

16 (2^{2 x}) + 15 (2^{x}) - 1 = 0

e^{x} - 5 e^{2 x} = 0

x, f = (\frac{1}{5})^{x}

3^{x + 2} + 3^{x - 1} = 84

9^{x} = 33

What is the answer to solvefor n,F=P(1+i)^n ?
The answer to solvefor n,F=P(1+i)^n is n=(ln(F/P))/(ln(1+i))