Question

If $3^{x+y}=81$ and ${81}^{x-y}=3$, then find the values of $x$ and $y$

Answer 1

$3^{x+y}=81$ and ${81}^{x-y}=3$

$\Rightarrow 3^{x+y}=3^4$ and $3^{4(x-y)}=3^1$

Equating the terms with the same bases, we have

$x+y=\mathrm{4......}\left(i\right)$

$4x-4y=\mathrm{1......}\left(ii\right)$

On multiplying $eq.\left(i\right)$ by $4$ and adding the result to eq.$\left(ii\right),$ we get

$\Rightarrow 4x-4y+(4x+4y)=1+16$

$\Rightarrow 8x=17$

$\therefore x=\frac{17}{8}$

Put $x=\frac{17}{8}$ in $eq.\left(i\right)$ we get

$\frac{17}{8}+y=4$

$\Rightarrow y=4-\frac{17}{8}=\frac{15}{8}$

$\therefore x=\frac{17}{8},y=\frac{15}{8}$