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 8 - 2017 - Paper 2

We start with the equality:

$gf(x) = fg(x)$

From part (a), we know that:

$gf(x) = 3x$

Now we need to calculate $fg(x)$:

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

Using the property of exponents:

$e^{3 ext{ln} x} = x^3$

Setting the two expressions equal gives:

$3x = x^3$

Rearranging this equation results in:

$x^3 - 3x = 0$

Factoring out an $x$ we have:

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

This leads to:

$x = 0 \, ext{or} \, x^2 - 3 = 0$

The second equation simplifies to:

$x = ext{±} ext{√3}$

Thus, the real solutions are $x = 0$, $x = ext{√3}$, and $x = - ext{√3}$. However, since $g(x)$ is only defined for $x > 0$, we only consider:

$x = ext{√3}$

Hence, the only real value of $x$ for which $gf(x) = fg(x)$ is:

$x = ext{√3}$