# Solve $6^x=36\cdot (9.75)^{x-2}$

I try to solve the following equation:

$6^x=36\cdot (9.75)^{x-2}$

I tried:

$1=6\cdot (9.75)^{x-2}$

But this is obviously wrong!

I think it would be smarter to bring the whole expression on one side. How should I proceed instead?

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Try $\log$ on both sides –  Shuchang Dec 26 '13 at 15:30
Have you $6^x=36.9,75^{x-2}$ or this equation: $6^x=36.9.75^{x-2}$? –  Marek Dec 26 '13 at 16:02
Hint: rewrite it as $\,(6/9.75)^{x-2} = 1.\ \$ –  Bill Dubuque Dec 26 '13 at 16:04

Taking log

$x\log 6 = \log 36 + (x-2)\log9.75$

$x(\log6 - \log9.75) = \log36 - 2\log9.75$

$x = \frac{\log36 - 2\log9.75}{\log6 - \log9.75}$

$x = 2\frac{\log6 - \log9.75}{\log6 - \log9.75}$

$x = 2$

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Divide each side by $36$ to get $6^{x-2}=9.75^{x-2}$. Hence $x=2$.
$$6^x=36\cdot9.75^{x-2}6^x=6^2(9+3/4)^{x-2}$$ $$6^{x-2}=(39/4)^{x-2}$$ $$\left(\frac{6}{39/4}\right)^{x-2}=1$$ $$\left(\frac{24}{39}\right)^{x-2}=\left(\frac{24}{39}\right)^0$$ $$x-2=0$$ $$x=2$$