# Solving this Logarithmic equation

• Nov 18th 2013, 03:20 AM
mohamedennahdi
Solving 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 AM
topsquark
Re: Solving this Logarithmic equation
Quote:

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

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

Hint: Start by taking log base 3 of both sides.

-Dan
• Nov 18th 2013, 11:34 AM
mohamedennahdi
Re: 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 PM
topsquark
Re: Solving this Logarithmic equation
Quote:

Originally Posted by mohamedennahdi
What do you think of this?

$\displaystyle \\3^{log_2(n)} = 3^{\frac{log_3(n)}{log_3(2)}}\\\\3^{log_2(n)} = n^{\frac{1}{log_3(2)}}\\\\3^{log_2(n)} = \sqrt[log_3(2)]{n}$

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 PM
mohamedennahdi
Re: Solving this Logarithmic equation
• Nov 18th 2013, 12:42 PM
topsquark
Re: Solving this Logarithmic equation
Quote:

Originally Posted by mohamedennahdi

Aw c'mon! Are you telling me you can't solve an equation of the form a*log_2(n) = b? Give it a try! I'll even tell you that log_3(3) = 1.

-Dan
• Nov 18th 2013, 01:02 PM
mohamedennahdi
Re: Solving this Logarithmic equation

$\displaystyle \\log_2(n).log_3(3) = log_3(x)\\\\log_2(n) = log_3(x)\\\\n = log_3(x)^2\\\\n = 2log_3(x)$
• Nov 18th 2013, 01:10 PM
topsquark
Re: Solving this Logarithmic equation
Quote:

Originally Posted by mohamedennahdi
$\displaystyle \\log_2(n).log_3(3) = log_3(x)\\\\log_2(n) = log_3(x)\\\\n = log_3(x)^2\\\\n = 2log_3(x)$
$\displaystyle log_2(n) = log_3(x)$
$\displaystyle 2^{log_2(n)} = 2^{log_3(x)}$
$\displaystyle n = 2^{log_3(x)}$