I would like to know how much 'n' equals, based on the formula below:

$\displaystyle 3^{log_2(n)} = X$

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- Nov 18th 2013, 03:20 AMmohamedennahdiSolving this Logarithmic equation
I would like to know how much 'n' equals, based on the formula below:

$\displaystyle 3^{log_2(n)} = X$ - Nov 18th 2013, 05:24 AMtopsquarkRe: Solving this Logarithmic equation
- Nov 18th 2013, 11:34 AMmohamedennahdiRe: Solving this Logarithmic equation
What do you think of this?

$\displaystyle \\3^{log_2(n)} = 3^{\frac{log_3(n)}{log_3(2)}}\\\\3^{log_2(n)} = n^{\frac{1}{log_3(2)}}\\\\3^{log_2(n)} = \sqrt[log_3(2)]{n}$ - Nov 18th 2013, 12:23 PMtopsquarkRe: Solving this Logarithmic equation
You still haven't separated the n from one side.

Here's where I was pointing you:

$\displaystyle 3^{log_2(n)} = x$

$\displaystyle log_3 \left ( 3^{log_2(n)} \right ) = log_3(x)$

$\displaystyle log_2(n) \cdot log_3(3) = log_3(x)$

Can you finish from here?

-Dan - Nov 18th 2013, 12:36 PMmohamedennahdiRe: Solving this Logarithmic equation
No, please proceed!

- Nov 18th 2013, 12:42 PMtopsquarkRe: Solving this Logarithmic equation
- Nov 18th 2013, 01:02 PMmohamedennahdiRe: Solving this Logarithmic equation
How about this:

$\displaystyle \\log_2(n).log_3(3) = log_3(x)\\\\log_2(n) = log_3(x)\\\\n = log_3(x)^2\\\\n = 2log_3(x)$ - Nov 18th 2013, 01:10 PMtopsquarkRe: Solving this Logarithmic equation