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?
$$3^{\frac{x-1}{2}}-2^{\frac{x+1}{3}}=2^{\frac{x-2}{3}}+3^{\frac{x-3}{2}}$$
 A: 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 $x=5$?
A: Bring same exponents to one side:
$$3^{\frac{x-1}{2}}-3^{\frac{x-3}{2}}=2^{\frac{x+1}{3}}+2^{\frac{x-2}{3}}$$
Factor out $3^{\frac{x-3}2}$ on the LHS and $2^{\frac{x-2}3}$ on the RHS:
$$3^{\frac{x-3}2}(3-1)=2^{\frac{x-2}3}(2+1)$$
Bring the $2$ and $3$ over again:
$$3^{\frac{x-3}2}\cdot2=2^{\frac{x-2}3}\cdot3$$
$$3^{\frac{x-5}2}=2^{\frac{x-5}3}$$
Take sixth powers, for neatness:
$$3^{3(x-5)}=2^{2(x-5)}$$
$$27^{x-5}=4^{x-5}$$
But the only place where graphs of two different exponential functions $a^x$ and $b^x$ intersect is at $x=0$. Hence, $x-5=0$ and $x=5$.
