What is the answer to solvefor i,\sqrt[2^i]{n}=2 ?

The answer to solvefor i,\sqrt[2^i]{n}=2 is i=-(ln(\frac{ln(2))/(ln(n)))}{ln(2)} {n>0}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor i,\sqrt[2^i]{n}=2

solve for

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

Solution steps

Apply exponent rules

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

Solve

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

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

Verify Solutions:

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

The solution is

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

22 = e^{x - 3} + e^{- (x - 3)}

4665 6^{2 x - 6} = 3 6^{x - 12}

8 = 10 (1 - e^{\frac{- t}{0.1}})

6^{2 x} = 25

4^{(2 x - 5)} = 64

What is the answer to solvefor i,\sqrt[2^i]{n}=2 ?
The answer to solvefor i,\sqrt[2^i]{n}=2 is i=-(ln(\frac{ln(2))/(ln(n)))}{ln(2)} {n>0}