<s>[INST] Exponential equations, solve for x I'm preparing for uni entrance exam. I've been struggling with this problem for about 90 minutes, tried everything I could think of. Can anybody explain how to solve for x step by step?
$$3^{\frac{x-1}{2}}-2^{\frac{x+1}{3}}=2^{\frac{x-2}{3}}+3^{\frac{x-3}{2}}$$
 [/INST] Hint: Your equation can be written as: $$3^{\frac{x-1}{2}} - 3^{\frac{x-3}{2}} = 2^{\frac{x-2}{3}} + 2^{\frac{x+1}{3}}.$$ Because $$3^{\frac{x-1}{2}} - 3^{\frac{x-3}{2}} = 3^{\frac{x-3+2}{2}} - 3^{\frac{x-3}{2}} = 3^{\frac{x-3}{2} +\frac{2}{2}} - 3^{\frac{x-3}{2}} = 3^{\frac{x-3}{2} +1} - 3^{\frac{x-3}{2}} = 3^{\frac{x-3}{2}}3^1 - 3^{\frac{x-3}{2}} = 3^{\frac{x-3}{2}}(3-1),$$ and 
$$2^{\frac{x-2}{3}} + 2^{\frac{x+1}{3}} = 2^{\frac{x-2}{3}} + 2^{\frac{x-2+3}{3}}= 2^{\frac{x-2}{3}} + 2^{\frac{x-2}{3}+\frac{3}{3}}=2^{\frac{x-2}{3}} + 2^{\frac{x-2}{3}+1}=2^{\frac{x-2}{3}} + 2^{\frac{x-2}{3}}2^1 = 2^{\frac{x-2}{3}}(1+2),$$
we can rewrite the first equation above as $$3^{\frac{x-3}{2}}(3-1) = 2^{\frac{x-2}{3}}(1+2).$$ Simplifying, we get $$3^{\frac{x-5}{2}} = 2^{\frac{x-5}{3}}.$$ Can you see why this implies $x=5$?
</s>