# Thread: Solve equation involving logarithms

1. ## 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!

2. 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!
$\dfrac{2^{4x}}{3^{3x}} = \dfrac{16^x}{27^x} = \left(\dfrac{16}{27} \right)^x = \dfrac{1}{2}$
Now use logarithms to the base $\frac{16}{27}$