Question

solve this friends...
${2}^{2x} - {2}^{x + 3} + 16 = 0\p$
Find the value of ’x’ ?
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Answer 1

${2}^{2x} - {2}^{x + 3} + 16 = 0 \\ {2}^{2x} - {2}^{x + 3} + 2 \times 2 \times 2 \times 2 = 0 \\ {2}^{2x } - {2}^{x + 3} + {2}^{4} =0 \\ on \: comparing \: the \: powers \\ 2x - (x + 3) + 4 = 0 \\ 2x - x - 3 + 4 = 0 \\ x + 1 = 0 \\ x = 1$

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