What is the answer to solvefor i, n/(2^i)=1 ?

The answer to solvefor i, n/(2^i)=1 is i=(ln(n))/(ln(2)) {n>0}

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solvefor i, n/(2^i)=1

solve for

i, \frac{n}{2 ^{i}} = 1

i = \frac{ln ( n )}{ln ( 2 )} {n > 0}

Solution steps

\frac{n}{2 ^{i}} = 1

Apply exponent rules

- i ln (2) = ln (\frac{1}{n})

Solve

- i ln (2) = ln (\frac{1}{n}) : i = \frac{ln ( n )}{ln ( 2 )}

i = \frac{ln ( n )}{ln ( 2 )}

Verify Solutions:

i = \frac{ln ( n )}{ln ( 2 )} {n > 0}

The solution is

i = \frac{ln ( n )}{ln ( 2 )} {n > 0}

x, y - 3 = 4 (5^{2 x + 1})

57 = 1 3^{x}

2^{6 \cdot l o g_{4} (x)} = 8

9^{x + 12} = 8 1^{x + 5}

e^{3 - 4 x} = 8

What is the answer to solvefor i, n/(2^i)=1 ?
The answer to solvefor i, n/(2^i)=1 is i=(ln(n))/(ln(2)) {n>0}