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

To show that there is only one real value of $x$ such that $gf(x) = fg(x)$, we first evaluate $fg(x)$:

$fg(x) = f(g(x)) = f(3 \ln x)$

Substituting into $f(x)$ gives us:

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

Next, we set $gf(x)$ equal to $fg(x)$:

$3x = x^3$

Rearranging gives:

$x^3 - 3x = 0$

Factoring this yields:

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

The solutions to this equation are:

$x = 0 \quad \text{or} \quad x^2 = 3$

Thus,

$x = 0 \quad \text{or} \quad x = \sqrt{3} \quad \text{or} \quad x = -\sqrt{3}$

However, since $g(x)$ is only defined for $x > 0$, we discard $x = 0$ and $x = -\sqrt{3}$. Therefore, the only valid solution is:

$x = \sqrt{3}$