Review question

# Which is the greater of $e^{\pi}$ and $\pi^e$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8671

## Solution

Show that for $x>0$, $x^{1/x}$ has its greatest value when $x=e$.

Let’s write $y=x^{1/x}$, and take the natural logarithm of both sides, giving $\ln y=\frac{1}{x} \ln x.$ Differentiating this with respect to $x$ using the product and chain rules gives \begin{align*} \frac{1}{y} \frac{dy}{dx}&=-\frac{1}{x^2}\ln x+\frac{1}{x^2}\\ &=\frac{1-\ln x}{x^2} \end{align*}

so $\frac{dy}{dx}=\frac{y(1-\ln x)}{x^2}=\frac{x^{1/x}(1-\ln x)}{x^2}.$ Turning points occur when $\dfrac{dy}{dx}=0$, which is precisely when $1-\ln x=0,$ as $x^{1/x}$ is never zero for $x>0$. So $\ln x=1$, and the only stationary point is at $x=e$.

Alternatively we can say, let $x^{1/x} = e^{f(x)}.$ Taking natural logs shows that $f(x) = \dfrac{\ln x}{x}$.

We can now differentiate $y = e^{f(x)}$ using the chain and quotient rules.

To determine the nature of the stationary point, we’ll look at the value of $y$ or $\dfrac{dy}{dx}$ either side of $x=e$.

We have $\dfrac{dy}{dx}=\dfrac{x^{1/x}(1-\ln x)}{x^2},$ and since $x^{1/x}$ and $x^2$ are always positive for $x>0$, the sign of the derivative is determined by the sign of $1-\ln x$.

So when $x<e$, $\dfrac{dy}{dx}>0$ and the function is increasing; when $x>e$, $\dfrac{dy}{dx}<0$ and the function is decreasing.

Therefore the function has a maximum at $x=e$, and this is the global maximum of the function.

Hence, or otherwise, but without use of a calculator, determine which is the greater of $e^{\pi}$ and $\pi^e$.

From our argument above, we must have $\pi^{1/\pi} < e^{1/e}$.

Raising both sides to the power $e \pi$ gives $\pi^e<e^{\pi}$ (this step is valid, since the function $f(x)=x^{e \pi}$ is strictly increasing for $x > 0$).

So $e^{\pi}$ is larger.