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How can I solve $$8(9^x)+3(6^x)-81(4^x)=0 $$ for $x$ using elementary methods.

Thanks a lot!

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    $\begingroup$ I see you added back Thanks a lot! which I had edited out. Just so you know, the Stack Exchange has decided that you don't need to add this (you can thank people by choosing their answer or voting for it). $\endgroup$ – Garrett Jun 21 '14 at 9:59
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Let's rewrite your expression a bit to make things clearer:

$$8\cdot 9^x + 3\cdot 6^x - 81\cdot4^x = 8\cdot (3^x)^2 + 3\cdot 3^x\cdot 2^x - 81 \cdot (2^x)^2.$$

If you want, set $y = 3^x$ and $z = 2^x$ to get

$$8y^2 + 3yz -81z^2 = 0.$$

Can you solve (i.e. factor) this? (Try quadratic equation.)

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Building on @Cameron Williams' method, we divide the original equation by $4^x$ and obtain:

$$8u^2 + 3u -81 = 0$$ where $$u=\left(\frac{3}{2}\right)^x$$

Once you solve it for $u$ then, $x=\frac{\ln u}{\ln(3/2)}$

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    $\begingroup$ Very clever, Mike! +1 $\endgroup$ – Cameron Williams Jun 20 '14 at 1:00

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