Asked by stay safe
solve and give answers in an exact form
(16^x)[1/(2^(x+2))] = 8[(4^x)]^-1
(16^x)[1/(2^(x+2))] = 8[(4^x)]^-1
Answers
Answered by
oobleck
(16^x)[1/(2^(x+2))] = 8[(4^x)]^-1
2^(4x) * 2^(-(x+2)) = 2^3 * 2^(-2x)
4x-x-2 = 3-2x
5x = 5
x = 1
check: (16^1)[1/(2^(1+2))] = 8[(4^1)]^-1
16/8 = 8/4
2 = 2
2^(4x) * 2^(-(x+2)) = 2^3 * 2^(-2x)
4x-x-2 = 3-2x
5x = 5
x = 1
check: (16^1)[1/(2^(1+2))] = 8[(4^1)]^-1
16/8 = 8/4
2 = 2
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