Question

Can some one help me. Solve this equation.

Answer 1

$\bold{Given : 2^2^x^-^2 - 2^x^-^1 = 2^x - 2}$

$\bold{\implies (2^2^x) (2^-^2) - (2^x)(2^-^1) = 2^x - 2}$

$\bold{\implies \frac{(2^2^x)}{(2^2)} - \frac{(2^x)}{(2^1)} = 2^x - 2}$

$\bold{\implies \frac{(2^2^x)}{4} - \frac{(2^x)}{2} = 2^x - 2}$

$\bold{\implies \frac{[2^2^x - (2^x)(2)]}{4} = 2^x - 2}$

$\bold{\implies [2^2^x - (2^x)(2)] = (4)(2^x) - 8}$

$\bold{\implies 2^2^x - (2^x)(2) - (4)(2^x) + 8 = 0}$

$\bold{\implies (2^x)^2 - (2^x)(6) + 8 = 0}$

$\bold{Let\;us\;take : (2^x) = Z}$

$\bold{\implies Z^2 - 6Z + 8 = 0}$

$\bold{\implies Z^2 - 4Z - 2Z + 8 = 0}$

$\bold{\implies Z(Z - 4) - 2(Z - 4) = 0}$

$\bold{\implies Z = 4\;\;(or)\;\;Z = 2}$

$\bold{But : Z = 2^x}$

$\bold{\implies 2^x = 4\;\; (or) \;\; 2^x = 2}$

$\bold{\implies 2^x = 2^2\;\; (or) \;\; 2^x = 2^1}$

$\bold{\implies x = 2\;\;(or)\;\; x = 1}$