Question

If ${\left(3\right)}^{x+y}=81and{\left(81\right)}^{x-y}=3$, find the values of $xandy$

Answer 1

Step 1: Simplify the given expression

$\begin{array}{l}{\left(3\right)}^{x+y}=81\\ \Rightarrow 3^{x+y}={\left(3\right)}^4\\ \Rightarrow x+y=4\mathrm{......}\left(1\right)\left(\text{Base are same so, equating powers}\right)\\ {\left(81\right)}^{x-y}=3\\ \Rightarrow {\left(3^4\right)}^{x-y}=3\\ \Rightarrow \left(3^{4\left(x-y\right)}\right)=3^1\left(\because {\left(a^m\right)}^n=a^{mn}\right)\\ \Rightarrow 4\left(x-y\right)=1\mathrm{......}\left(2\right)\left(\text{Base are same so, equating powers}\right)\\ \end{array}$

Step 2: Compare both the equations

$\begin{array}{l}x+y=4\\ 4x-4y=1....\left(4\right)\\ \Rightarrow x=4-y\mathrm{......}\left(3\right)\left(\text{from equation 1}\right)\\ \Rightarrow 4\left(4-y\right)-4y=1(x\text{value replace in}(4\left)\right)\\ \Rightarrow 16-4y-4y=1\\ \Rightarrow -8y=1-16\\ \Rightarrow -8y=-15\\ \Rightarrow y=\frac{15}{8}\end{array}$

Step 3: Substitute the value of $yineq.\left(3\right)$

$\begin{array}{l}x=4-y\\ \Rightarrow x=4-\frac{15}{8}\\ \Rightarrow x=\frac{32-15}{8}\\ \Rightarrow x=\frac{17}{8}\end{array}$

Therefore , the value of $xandy$ are $\frac{17}{8}and\frac{15}{8}$respectively.