Question

Solve the following systems of equations.
$3^{2x}-2^y=77,3^x-2^{y/2}=7$

Answer 1

Let $3^{2x}-2^y=77...\left(1\right)$

$3^x-2^{y/2}=7...\left(2\right)$

by eq (2) $3^x=7+2^{y/2}$

taking square both sides

$3^{2x}=(7+2^{y/2}{)}^2$

putting value of $3^{2x}$ in eq(1)

$(7+2^{y/2}{)}^2-2^y=77$

$49+{14.2}^{y/2}+2^y-2^y=77$

$2^{y/2}=2$

taking log both sides

$y/2log2=log2$

$y=2$

putting value of $y=2$ in eq.(1)

$3^{2x}-2^2=77$

$3^{2x}=81$

$3^{2x}=3^4$

taking log both sides

$2xlog3=4log3$

$x=2$

sol.$x=2$

$y=2$