Solve the equation $9^x+3^{2x-1}=2^{x+3.5}+2^{x+0.5}$ Solve the equation $$9^x+3^{2x-1}=2^{x+3.5}+2^{x+0.5}$$
The equation is equivalent to $$3^{2x}+3^{2x}\cdot3^{-1}=2^x\cdot2^{3.5}+2^x\cdot2^{0.5}$$ which is $$3^{2x}\left(1+\dfrac{1}{3}\right)=2^x\left(2^\frac72+2^\frac12\right)$$ We can write the last equation as $$\dfrac433^{2x}=9\sqrt2\cdot2^x,$$ or $$\dfrac{3^{2x}}{2^x}=\dfrac{9\sqrt2}{\frac43}\iff\left(\dfrac{9}{2}\right)^x=\dfrac{3^3}{2^{2-\frac12}}=\dfrac{3^3}{2^\frac32}$$ I don't see where my solution goes wrong as the answer is $x=\dfrac{3}{2}=1.5$. Any thoughts on that?

you just need to simplify the last expression as
$\displaystyle \left(\frac{9}{2}\right)^{x} \ =\ \frac{27\sqrt{2}}{4} \ =\ \left(\frac{3^{3}}{2^{( 3/2)}}\right) \ =\ \left(\frac{9^{1.5}}{2^{1.5}}\right)$
