Solve $2(4^x) + 3(9^x)=5(6^x)$

Question

Ended up with $2^{2x+1} + 3^{2x+1} = 5(6^x)$. Not sure how to proceed from here. Thanks!

Answer 1

I would write this as $$2a^2-5ab+3b^2=0\tag1$$ with $a=2^x$ and $b=3^x$ and hope I could factor (1).

Answer 2

Let $u = 2^x$, and $v = 3^x$, so $4^x = u^2, 9^x = v^2, 6^x = uv$

The equation you have is $2 u^2 + 3 v^2 = 5 uv$

Factor this as $(2u-3v)(u-v) = 0$

Thus, $2u = 3v$ or $u = v$. If $u = v$, then we get $2^x = 3^x$. Taking logs by 2, we get $x = x \log_2(3)$. Then, as $1 \neq \log_2(3)$, we must have $x = 0$.

In the second case, we get $2u = 3v$, or $2 * 2^x = 3 * 3^x$. That is, $2^{x+1} = 3^{x+1}$. Again, taking logs and following the idea above, we get $x = -1$.

The solutions are $x = 0, x = -1$.

Answer 3

Hint: put $y=2^x, z=3^x$ and factorise the resulting quadratic.

Answer 4

Hint: $\displaystyle\;\;2\cdot\left(\frac{2}{3}\right)^x + 3 \cdot \left(\frac{3}{2}\right)^x$ is strictly convex on $\mathbb{R}\,$, and equals $\,5\,$ at $\,x=0, -1\,$.