Question

koi mera help krega ??
and please don't put wrong answer​

Answer 1

Step-by-step explanation:

i think it will be the ansr

Answer 2

Step-by-step explanation:

We have,

${2}^{2x} - 3 \times {2}^{x + 2} + 32 = 0$

$\implies ({2}^{x} )^{2} - 12. {2}^{x} + 32 = 0$

$let \: \: {2}^{x} = y$

Then, we have,

${y}^{2} - 12y + 32 = 0$

$\implies {y}^{2} - 4y - 8y + 32 = 0$

$\implies(y - 4)(y - 8) = 0$

$\implies \: y = 4 \: \: or \: \: y = 8$

$\implies {2}^{x} = 4 \: \: or \: \: {2}^{x} = 8$

$\implies \: x = 2 \: \: or \: \: x = 3$