# Solving this Logarithmic equation

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• Nov 18th 2013, 04:20 AM
mohamedennahdi
Solving this Logarithmic equation
I would like to know how much 'n' equals, based on the formula below:

$3^{log_2(n)} = X$
• Nov 18th 2013, 06:24 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:

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

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

-Dan
• Nov 18th 2013, 12:34 PM
mohamedennahdi
Re: Solving this Logarithmic equation
What do you think of this?

$\\3^{log_2(n)} = 3^{\frac{log_3(n)}{log_3(2)}}\\\\3^{log_2(n)} = n^{\frac{1}{log_3(2)}}\\\\3^{log_2(n)} = \sqrt[log_3(2)]{n}$
• Nov 18th 2013, 01:23 PM
topsquark
Re: Solving this Logarithmic equation
Quote:

Originally Posted by mohamedennahdi
What do you think of this?

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

You still haven't separated the n from one side.

Here's where I was pointing you:
$3^{log_2(n)} = x$

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

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

Can you finish from here?

-Dan
• Nov 18th 2013, 01:36 PM
mohamedennahdi
Re: Solving this Logarithmic equation
No, please proceed!
• Nov 18th 2013, 01:42 PM
topsquark
Re: Solving this Logarithmic equation
Quote:

Originally Posted by mohamedennahdi
No, please proceed!

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, 02:02 PM
mohamedennahdi
Re: Solving this Logarithmic equation
How about this:

$\\log_2(n).log_3(3) = log_3(x)\\\\log_2(n) = log_3(x)\\\\n = log_3(x)^2\\\\n = 2log_3(x)$
• Nov 18th 2013, 02:10 PM
topsquark
Re: Solving this Logarithmic equation
Quote:

Originally Posted by mohamedennahdi
How about this:

$\\log_2(n).log_3(3) = log_3(x)\\\\log_2(n) = log_3(x)\\\\n = log_3(x)^2\\\\n = 2log_3(x)$

To get rid of a "log" you have to raise it to a power. So after line 2 we would have
$log_2(n) = log_3(x)$

$2^{log_2(n)} = 2^{log_3(x)}$

$n = 2^{log_3(x)}$

-Dan