Solve for x....

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

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- Nov 6th 2009, 07:30 AMCantcountWontcountLogarithms... I think?
Solve for x....

2^(2x+1) = 3^(2x-5) - Nov 6th 2009, 07:40 AMmathaddict
HI

Apply log to both sides ,

$\displaystyle

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

$

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

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

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

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

Can you take it from here ? - Nov 6th 2009, 07:51 AMCantcountWontcount
Thanks addict, the answer on my tutorial sheet must be wrong...

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