# Logarithms... I think?

• November 6th 2009, 07:30 AM
CantcountWontcount
Logarithms... I think?
Solve for x....

2^(2x+1) = 3^(2x-5)
• November 6th 2009, 07:40 AM
Quote:

Originally Posted by CantcountWontcount
Solve for x....

2^(2x+1) = 3^(2x-5)

HI

Apply log to both sides ,

$
\log 2^{2x+1}=\log 3^{2x-5}
$

$(2x+1)\log 2=(2x-5)\log 3$

$\frac{2x+1}{2x-5}=\frac{\log 3}{\log 2}$

$\frac{2x+1}{2x-5}=1.585$

$2x+1=1.585(2x-5)$

Can you take it from here ?
• November 6th 2009, 07:51 AM
CantcountWontcount

I completed it by doing

(2x + 1)log2=(2x - 5)log3
2xlog2+log2=2xlog3-5log3
2xlog3-2xlog2=log2+5log3
2x(log3-log2)=log2+5log3
2x=(log2+5log3)/(log3-log2)......

I get the same answer doing it this way, was pulling my hair out for ages.

Thanks again