Solve equation involving logarithms

**mathguy80** · April 5th 2011, 11:01 PM

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!

**earboth** · April 5th 2011, 11:08 PM

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 $\frac{16}{27}$

**mathguy80** · April 6th 2011, 12:00 AM

Thanks @earboth. Looks simple now!

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