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

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

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

solve for

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

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

Solution steps

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

Apply exponent rules

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

Solve

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

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

Verify Solutions:

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

The solution is

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

t, e^{k t} = \frac{1}{2}

2^{x - 3} = \frac{1}{32}

e^{4 x} = 9

8^{2 x + 1} = 1

2^{x - 2} = 16

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