Given f(x) = 2^(log₂ x) + 8 (with x 0), find the inverse function f⁻¹(x) and hence solve f(x) = f⁻¹(x).

Question

Given f(x) = 2^(log₂ x) + 8 (with x > 0), find the inverse function f⁻¹(x) and hence solve f(x) = f⁻¹(x).

Accepted Answer

f⁻¹(x) = x - 8; no real solution. Since 2^(log₂ x) = x (for x > 0), f(x) = x + 8. Its inverse is f⁻¹(x) = x - 8. Setting f(x) = f⁻¹(x): x + 8 = x - 8 gives 8 = -8, which is impossible, so there is no solution. (Both f and f⁻¹ are parallel lines of slope 1 offset by 16, so they never meet.)

Given f(x) = 2^(log₂ x) + 8 (with x > 0), find the inverse function f⁻¹(x) and hence solve f(x) = f⁻¹(x).

Solution

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