Question

Solve for x and y, if (√32)^x÷2^y+1=1 and 8^y-16^y-x/2=0

Answer 1

Answer:

Quite easy.

$( { \sqrt{32} })^{x} = {2}^{ \frac{5x}{2} }$

so,

${2}^{ \frac{5x}{2} } \div {2}^{y + 1} = 1$

$\frac{5x}{2} - (y + 1) = 0$

and

$3y = 4(4 - \frac{x}{2})$

Solving we get,

$x = 2$

$y = 4$