Logarithms... I think?

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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
mathaddict

Quote:

Originally Posted by CantcountWontcount

Solve for x....

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

HI

Apply log to both sides ,

$<br /> \log 2^{2x+1}=\log 3^{2x-5}<br />$

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

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

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

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

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

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

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