Solve for x to eight decimal places if

2^(2x – 1) = 3^(x-2)

Substitute your x value answer

back into original and show that it

will make each side equal.

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- Nov 6th 2008, 05:40 PMAlesa'7All help is GREATLY appreciated!!! :)
Solve for x to eight decimal places if

2^(2x – 1) = 3^(x-2)

Substitute your x value answer

back into original and show that it

will make each side equal. - Nov 6th 2008, 05:42 PMvincisonfire
$\displaystyle 2(2x – 1) = 3(x-2)$

$\displaystyle 4x – 2 = 3x-6$

$\displaystyle x = -4$ - Nov 6th 2008, 05:52 PMAlesa'7
Oh I am sorry those numbers in parenthesis are suppost to be exponets (2x-1) and (x-2) are exponets, i am so sry.. :(

- Nov 6th 2008, 06:24 PMmr fantastic
$\displaystyle 2^{2x-1} = 3^{x-2} \Rightarrow \log_{10} 2^{2x-1} = \log_{10} 3^{x-2} \Rightarrow (2x - 1) \log_{10} 2 = (x - 2) \log_{10} 3$.

This has the form $\displaystyle a(2x-1) = b(x - 2)$. Expand and make x the subject. Then use a calculator to get the required approximate value. - Nov 6th 2008, 06:40 PMAlesa'7Response to your comment.
yea, if you would have read then you would see that I had written it incorrectly , and I had already thanked him, and others who are Trying to Help, would see that it had already been replied to and then would move on and not bother.. Soo.. I rewrote the question showing that it hadnt in fact been answered at all, (correctly any way).as of now, by the way, it still hasnt been correctly solved. :)