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Let f: R -> R+ be defined by f(x) = e^x and g: R -> R be defined by g(x) = 30 - 2x. Let fₙ(x) denote the n-fold composition of f with itself (f applied n times). How many solutions does the equation fₙ(g(x)) = f_(n-1)(-x) have?

1. 0
2. 1
3. 2
4. infinite

Correct answer: 1

Solution

Since f(x) = e^x is injective, fₙ(g(x)) = f_(n-1)(-x) implies f_(n-1)(g(x)) = ln(f_(n-1)(-x)). Repeatedly applying the inverse (natural log) n-1 times reduces the equation to g(x) = -x, i.e., 30 - 2x = -x, giving x = 30. There is exactly 1 solution.

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