Given f(x) = e^x, x ∈ ℝ g(x) = 3 ln x, x > 0, x ∈ ℝ (a) find an expression for gf(x), simplifying your answer - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 2

We need to show that there is only one real value of x such that gf(x) = fg(x).

1. Start with the expressions obtained:

   $gf(x) = 3x$

   $fg(x) = f(g(x)) = f(3 ext{ln}x)$

2. Substitute into fg(x):

   $fg(x) = e^{(3 ext{ln}x)}$

3. Using the property $e^{ ext{ln}(a)} = a$, we get:

   $fg(x) = x^3$

4. Set the two expressions equal to each other:

   $3x = x^3$

5. Rearranging gives:

   $x^3 - 3x = 0$

6. Factor out x:

   $x(x^2 - 3) = 0$

7. This gives us two cases:

   • Case 1: $x = 0$
   • Case 2: $x^2 - 3 = 0 ightarrow x = ext{±} ext{√3}$

8. Since g(x) is defined only for x > 0, we disregard x = 0 and only consider $x = ext{√3}$.

Therefore, there is only one real value for which gf(x) = fg(x):

$x = ext{√3}$