Solve 8^x + 4 = 4^x + 2^(x+2).
Firstly, we want to notice that this question only involves powers of 2^x, ie.
(2^x)^3 + 4 = (2^x)^2 + 4(2^x)
so we can use the substitution u = 2^x to give
u^3 - u^2 - 4u + 4 = 0
The problem (at least at this intermediate stage) has reduced to solving a cubic equation in the variable u. u = 1 satisfies this equation, so by the factor theorem, we know that u - 1 is a factor. Then by equating coefficients or polynomial long division, we can write the equation as
(u - 1)(u^2 - 4) = 0
or equivalently
(u - 1)(u - 2)(u +2) = 0
Substituting back in for x, we have
u = 1 => 2^x = 1 => x = 0
u = 2 => 2^x = 2 => x = 1
u = -2 => 2^x = -2 which has no real solutions.
Hence, the solutions to the original equation are x = 0, 1.
SA
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