# Solve equation involving logarithms

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• Apr 6th 2011, 12:01 AM
mathguy80
Solve equation involving logarithms
Hi All,

I was studying logarithms today, and came across this problem. Need a hint on how to proceed.

$2^{4x} - 3^{3x} + 4^{2x} = 0$

I simplified it down to,

$\frac{2^{4x}}{3^{3x}} = \frac{1}{2}$

How do I proceed further? Thanks!
• Apr 6th 2011, 12:08 AM
earboth
Quote:

Originally Posted by mathguy80
Hi All,

I was studying logarithms today, and came across this problem. Need a hint on how to proceed.

$2^{4x} - 3^{3x} + 4^{2x} = 0$

I simplified it down to,

$\frac{2^{4x}}{3^{3x}} = \frac{1}{2}$

How do I proceed further? Thanks!

I take your last line:

$\dfrac{2^{4x}}{3^{3x}} = \dfrac{16^x}{27^x} = \left(\dfrac{16}{27} \right)^x = \dfrac{1}{2}$

Now use logarithms to the base $\frac{16}{27}$
• Apr 6th 2011, 01:00 AM
mathguy80
Thanks @earboth. Looks simple now!