What is the answer to solvefor n,s=r(((1+i)^n-1))/i ?

The answer to solvefor n,s=r(((1+i)^n-1))/i is n=(ln((si+r)/r))/(ln(1+i))

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solvefor n,s=r(((1+i)^n-1))/i

solve for

n, s = r \frac{( ( 1 + i ) ^{n} - 1 )}{i}

n = \frac{ln ( \frac{s i + r}{r} )}{ln ( 1 + i )}

Solution steps

s = r \frac{( ( 1 + i ) ^{n} - 1 )}{i}

Switch sides

r \frac{( 1 + i ) ^{n} - 1}{i} = s

Refine

r \frac{( 1 + i ) ^{n} - 1}{i} : \frac{r ( ( 1 + i ) ^{n} - 1 )}{i}

\frac{r ( ( 1 + i ) ^{n} - 1 )}{i} = s

Multiply both sides by

i

r ((1 + i)^{n} - 1) = s i

Expand

r ((1 + i)^{n} - 1) : r (1 + i)^{n} - r

r (1 + i)^{n} - r = s i

Move

r

to the right side

r (1 + i)^{n} = s i + r

Divide both sides by

r

(1 + i)^{n} = \frac{s i + r}{r}

Apply exponent rules

n ln (1 + i) = ln (\frac{s i + r}{r})

Solve

n ln (1 + i) = ln (\frac{s i + r}{r}) : n = \frac{ln ( \frac{s i + r}{r} )}{ln ( 1 + i )}

n = \frac{ln ( \frac{s i + r}{r} )}{ln ( 1 + i )}

9^{x} = 4

9^{x} = 9

2^{[x] + [x - 2]} = 16

2 5^{x - 1} + 100 = \frac{( 5 ^{2 x} )}{5}

a, a^{a} + 5 = b^{b} + a^{a} - 2^{2}

What is the answer to solvefor n,s=r(((1+i)^n-1))/i ?
The answer to solvefor n,s=r(((1+i)^n-1))/i is n=(ln((si+r)/r))/(ln(1+i))