Solve: $4(9^x)+3(4^x)=13(6^x)$

Question

I'm new to this site so bear with me if I make some mistakes. Here is the question:

Solve: $4(9^x)+3(4^x)=13(6^x)$

I have absolutely no clue how to attack this problem. I have tried a bit of logs but I come to a dead end. What could I do?

Answer 1

Dividing both sides by $4^{x}$:

$4(\frac{9}{4})^{x} + 3 = 13(\frac{3}{2})^{x}$

$4(\frac{3}{2})^{2x} - 13(\frac{3}{2})^{x} + 3 = 0$

Note that this is a quadratic in $(\frac{3}{2})^{x}$, which can be factored as:

$(4(\frac{3}{2})^{x}-1)((\frac{3}{2})^{x}-3) = 0$

Thus, we either have $(\frac{3}{2})^{x} = \frac{1}{4}$ or $(\frac{3}{2})^{x}=3$, which gives a solution set of $\boxed{x = -\log_{\frac{3}{2}}(4), \log_{\frac{3}{2}}(3).}$

If you want, you can use change-of-base and log rules to get everything in normal $\log$s.

Answer 2

Hint: Rewrite $4(9^x) + 3(4^x) = 13(6^x)$ as $4(3^x)^2 + 3(2^x)^2 = 13 (2^x)(3^x)$.