Consider this function, defined for every x > 1 (where "ln" indicates the natural logarithm):
f(x) = 2ln(x) − 2ln(x − 1)
Which of these is an expression for its inverse function f⁻¹(y)?
A) 1 / (e^(y/2) − 1), y > 0
B) 1 / (e^(y/2) − 1), y > 1
C) e^(y/2) / (e^(y/2) − 1), y > 0
D) 1 / (e^(y/2) − 1), y ≠ 0
E) e^(y/2) / (e^(y/2) − 1), y > 1
IMAT 2023 · Question 48
Q48 - Logarithms, composite and inverse functions
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The correct answer is C) e^(y/2) / (e^(y/2) − 1), y > 0.
Combine the logs first: y = 2 ln[x/(x − 1)], so x/(x − 1) = e^(y/2).
Write k = e^(y/2) and solve: x = k(x − 1) → x(k − 1) = k → x = e^(y/2) / (e^(y/2) − 1).
For the domain: when x > 1, x/(x − 1) is always bigger than 1, so its log is positive and y > 0. That gives C.
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