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 6 - 2017 - Paper 2

To show that there is only one real value of x for which gf(x) = fg(x), we first express fg(x):

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

Substituting gives:

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

Using the exponent property ( e^{\ln a} = a ), we have:

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

Now, we set the two expressions equal:

$gf(x) = fg(x) \Rightarrow 3x = x^3$

Rearranging gives:

$x^3 - 3x = 0$

Factoring out x gives:

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

This results in:

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

Solving for the second equation leads to:

$x = \pm\sqrt{3}$

Thus, we have three possible solutions: x = 0, x = \sqrt{3}, and x = -\sqrt{3}.

However, since g(x) is only defined for x > 0, the only valid solution is:

$x = \sqrt{3}$

Thus, we conclude that there is only one real value of x for which gf(x) = fg(x):

$x = \sqrt{3}$