Question

This is a question from indices chapter. Anyone plz solve it​

Answer 1

Answer:

hope you got your answer.

Answer 2

Answer:

x = 5

Step-by-step explanation:

$given \\ \\ {4}^{x - 2} - {2}^{x + 1} = 0 \\ \\ \frac{ {4}^{x} }{ {4}^{2} } - 2 \times {2}^{x} = 0 \\ \\ \frac{ { ({2}^{2}) }^{x} }{ {4}^{2} } - 2 \times {2}^{x} = 0 \\ \\ \frac{ {( {2}^{x} )}^{2} }{16} - 2 \times {2}^{x} = 0 \\ \\ {2}^{x} ( \frac{ {2}^{x} }{16} - 2) = 0 \\ \\ {2}^{x} = 0 \\ \\ {2}^{x} = {2}^{0} \\ \\ = > x = 0 \\ \\ \frac{ {2}^{x} }{16} - 2 = 0 \\ \\ = > {2}^{x} = 32 \\ \\ {2}^{x} = {2}^{5} \\ \\ = > x = 5 \\ \\ x = 0 \: \: \: does \: not \: satisfy \: the \: given \: eqn \\ \\ = > x = 5$

HOPE IT HELPS