Exponential Equation

Let $u=3^{\frac{x-10}{10}}=3^{\frac{x}{10}-1},$ and note that $3^{\frac{x}{5}} = (3u)^2 =9u^2.$ now we are left to solve the quadratic equation $9u^2+u-84=0,$ which has solutions $u=3, -\frac{28}{9}.$ As u is a power of 3, we can throw out the negative solution for u and solve $3^{\frac{x-10}{10}}=3,$ leading us to the unique solution of $x=20.$ So, $A=1, B=20 \implies A+B=21.$

$\begin{aligned} 3^{\frac{x}{5}} + 3^{\frac{x-10}{10}} & = 84 \\ 3^{\frac{x}{5}} + 3^{\frac{x}{10}-1} & = 84 \\ \left( \color{#3D99F6}{3^{\frac{x}{10}}}\right)^2 + \frac{1}{3}\left( \color{#3D99F6}{3^{\frac{x}{10}}}\right) & = 84 \quad \quad \small \color{#3D99F6}{\text{Let }y = 3^{\frac{x}{10}}} \\ \Rightarrow 3\color{#3D99F6}{y}^2 + \color{#3D99F6}{y} - 252 & = 0 \\ (3y+28)(y-9) & = 0 \\ \Rightarrow y & = \begin{cases} 3^{\frac{x}{10}} = - \frac{28}{3} & \color{#D61F06}{< 0 \text{ rejected}} \\ 3^{\frac{x}{10}} = 9 = 3^2 & \Rightarrow \dfrac{x}{10} = 2 & \Rightarrow x = 20 \end{cases} \end{aligned}$

Therefore, $A+B = 1+20 = \boxed{21}$

A simpler solution would be to cycle through the values of x until you get 84. We know that x has to be a positive multiple of 5 and that x-10 has to be a positive multiple of 10 in order for the result to be an integer, so to get 84, just set the value of x to 20.

So A=1 solution and B=20, giving us A+B=21.