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Rieksie
ANSWERS: 2
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The equation says, 2^2x + 2^(x+1) = 2^3 => (2^x)^2 + 2 * 2^x = 2^3 =8 [as, a^(m+n)= a^m * a^n] => (2^x)^2 +2 *2^x *1 +1^2 = 8+1^2=9 [adding 1^2 to both] => (2^x +1)^2 = 9= 3^2 => 2^x +1 = 3 [taking sq root] => 2^x = 2= 2^1 => x =1 [if a^m = a^n => m=n] This is the solution to your problem. Note that, we discarded the negative magnitude while taking sq root as it is meaningless to write 2^x = -4. Hope, this'll help you.
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http://www.wolframalpha.com/input/?i=+2^2x%2B2^%28x%2B1%29%3D2^3+
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