Solve the following equations: $2^{2x + 3} - 57 = 65 (2^{x} - 1)$.
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Solve the following equations: $2^{2x + 3} - 57 = 65 (2^{x} - 1)$.
Let y = 2^x. The equation is 2^3 * y^2 - 57 = 65(y - 1), which is 8y^2 - 65y + 8 = 0. Factoring: (8y - 1)(y - 8) = 0. So y = 1/8 or y = 8. 2^x = 2^-3 or 2^x = 2^3, so x = -3 or x = 3.
Rewrite 2^(2x + 3) as 8 * 2^(2x), and let t = 2^x to transform the equation into 8t^2 - 57 = 65(t - 1). Simplifying this yields 8t^2 - 65t + 8 = 0, which factors into (8t - 1)(t - 8) = 0, meaning t = 8 or t = 1/8. Setting 2^x = 8 gives x = 3, and setting 2^x = 1/8 gives x = -3, so the solutions are 3 and -3.