Solve the equation $2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$ Solve the equation $$2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$$
The given equation is equivalent to $$2^{2x}+\dfrac12\cdot2^{2x}=3^x\sqrt3+\dfrac{3^x}{\sqrt3}$$ which is $$\dfrac32\cdot2^{2x}=3^x\left(\sqrt3+\dfrac{1}{\sqrt3}\right)$$ The last equation can be written as $$\dfrac32\cdot2^{2x}=\dfrac{4}{\sqrt{3}}\cdot3^x,$$ or $$\dfrac{2^{2x}}{3^x}=\dfrac{\frac{4}{\sqrt3}}{\frac{3}{2}}\iff\left(\dfrac43\right)^x=\dfrac{\frac{4}{\sqrt3}}{\frac{3}{2}}$$ How do I find $x$ from here, as it is obviously a difficulty for me? What's the approach supposed to be?
 A: Alternative approach:

*

*Let $r = \log_2(3).$


*Note that $2^a = 2^b \iff a = b.$

Solve the equation $$2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$$

The LHS can be re-written as
$$2^{2x} \times \left[1 + \frac{1}{2}\right] = 2^{2x} \times \frac{3}{2}.$$
The RHS can be re-written as
$$3^{x + 0.5} \times \left[1 + \frac{1}{3}\right] = 3^{x + 0.5} \times \frac{4}{3}$$
$$ = \left(2^{r}\right)^{x+0.5} \times \frac{4}{3} = \left[2^{r \times (x+0.5)}\right] \times \frac{4}{3}.$$
Since the LHS and RHS are equivalent, you know that
$$2^{2x-3} = \frac{2^{2x}}{8} = \frac{\left[2^{r \times (x+0.5)}\right]}{9} = 2^{r \times (x + 0.5 -2)}.$$
Therefore,
$$2^{2x-3} = 2^{r \times \left[x - \frac{3}{2}\right]} \implies $$
$$2x - 3 = r \times \left[x - \frac{3}{2}\right] \implies $$
$$x(2 - r) = 3 - \left[r \times \frac{3}{2}\right] \implies $$
$$x = \frac{3 - \left[r \times \frac{3}{2}\right]}{2 - r} = \frac{3}{2}.$$
A: Since your last step then $(4/3)^{x}=(4/3)^{3/2}$ then $x=3/2$ as Sil noticed.
Alternatively, let $x$ be a real number, then
\begin{align*}
2^{2x}+2^{2x-1}&=3^{x+1/2}+3^{x-1/2}\\ 
\frac{1}{3^{x+1/2}}\left(2^{2x}+2^{2x-1} \right)&=\frac{1}{3^{x+1/2}}\left(3^{x+1/2}+3^{x-1/2} \right)\\
2^{2x}\cdot3^{-(x+1/2)}+2^{2x-1}\cdot 3^{-(x+1/2)}&=3^{x+1/2}\cdot 3^{-(x+1/2)}+3^{x-1/2}\cdot 3^{-(x+1/2)}\\
2^{2x}\cdot3^{-(x+1/2)}+2^{2x-1}\cdot 3^{-(x+1/2)}&=4/3\\
\underbrace{e^{\ln 2^{2x}}\cdot e^{\ln 3^{-(x+1/2)}}}_{(*)}+e^{\ln 2^{2x-1}}\cdot e^{\ln 3^{-(x+1/2)}}&=4/3
\end{align*}
For $(*)$ note that is just $e^{(2x)\ln 2-(x+1/2)\ln 3}:=t$, then the above equation came be written as $\frac{3t}{2}=\frac{4}{3}$, that is, $t=8/9$.
Now, we need to find $x$ into $8/9=e^{(2x)\ln 2-(x+1/2)\ln 3}$. Applying logarithm both sides we have  $(2x)\ln 2-(x+1/2)\ln 3=\ln(8/9)$. Rewriting in order to find $x$ we have $(2\ln 2-\ln 3)x=1/2\ln 3+\ln(8/9)$. Thus ,
$$\boxed{x}=\frac{\frac{\ln 3}{2}+\ln\frac{8}{9}}{2\ln 2-\ln 3}=\boxed{\frac{\frac{1}{2}\ln 3+3\ln 2-2\ln 3 }{2\ln 2 -\ln 3}=\frac{3}{2}}$$
