Solve equation with $x$ in the exponents

Question

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

Thanks a lot!

Answer 1

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.)

Answer 2

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, $x=\frac{\ln u}{\ln(3/2)}$