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

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

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

solve for

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

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

Solution steps

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

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}

4 e^{x} = 17

5^{x - 2} + 5^{x - 1} = \frac{30}{5}

0 = e^{- t}

3^{2 x + 5} = 3^{x - 1}

1 4^{x + 3} = 20

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