Question

If $8^x\times 4^y=32$ and ${81}^x÷{27}^y=3;$ find the values of x

Answer 1

For $8^x\times 4^y=32$
$\Rightarrow {\left(2^3\right)}^x\times {\left(2^2\right)}^y=2^5$
And for ${81}^x÷{27}^y=3;$
$\Rightarrow {\left(3^4\right)}^x÷{\left(3^3\right)}^y=3^1$
$\therefore 2^{3x}\times 2^{2y}=2^5$ and $3^{4x}÷3^{3y}=3^1$
Now by using laws of indices $a^m\times a^n=a^{m+n}$ and $\frac{a^m}{a^n}=a^{m-n}$
$2^{3x+2y}=2^5$ and $3^{4x-3y}=3^1$
As base are same, we can equate the powers
So we have two equation
$\Rightarrow 3x+2y=5eq1$
$\Rightarrow 4x-3y=1eq2$
Now multiplying eq1 from 3 and Eq2 from 2 and adding both the equation
$\Rightarrow 9x+6y+8x-6y=15+2$
$\Rightarrow 17x=17$
$\Rightarrow x=1$
Now putting x = 1 in eq1
$\Rightarrow 3\left(1\right)+2y=5$
$\Rightarrow 2y=5-3$
$\Rightarrow 2y=2$
$\Rightarrow y=1$
Hence $x=1,y=1$