Can some one help me. Solve this equation.

Question 1

Question 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Please log in or register to add a comment.