<s>[INST] 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?
 [/INST] 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,
$$2^{2x-3} = 2^{r \times \left[x - \frac{3}{2}\right]} \implies $$
$$2x - 3 = r \times \left[x - \frac{3}{2}\right] \implies $$
$$x(2 - r) = 3 - \left[r \times \frac{3}{2}\right] \implies $$
$$x = \frac{3 - \left[r \times \frac{3}{2}\right]}{2 - r} = \frac{3}{2}.$$
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