Eleanor from English Martyrs Hartlepool and Jaro, Giulia and Pasan from St Louis School of Milan have tackled the problem of which is bigger out of $2^x$ and $x^2$.

Eleanor has started by considering a few specific values of $x$ to get an idea for what happens.

$2^0 > 0^2$    $1>0$                    

$2^1 > 1^2$    $2>1$

$2^2 = 2^2$    $4=4$

$2^3 < 3^2$    $8<9$

$2^4 = 4^2$    $16=16$

The main thing we notice here is that the answer depends on $x$.

Jaro, Giulian and Pasan have considered how the answer depends on $x$ more generally by considering the graphs of $2^x$ and $x^2$

To find out which one was bigger we drew a graph on which we could clearly

see the point of intersections of the two functions, of which there were 3, when

$x=-0.766, 2 \text{ and } 4$. Then, we reached the conclusion that:

- when $x$ is bigger than $4$, $2^x$ is bigger than $x^2$

- when $x$ is smaller than $4$ but bigger than $2$, $2^x$ is smaller than $x^2$

- when $x$ is smaller than $2$ but bigger than $-0.766$, $2^x$  is bigger than $x^2$

- when $x$ is smaller than $-0.766$, $2^x$ is smaller than $x^2$

 

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They also compared the graphs $3^x$ and $x^3$ as a first step to considering the general case of comparing $a^x$ and $x^a$ and found that there were again two positive intersections, this time at $x=2.48 \text{ and }3$.

Pablo and Sergio from King College of Alicante and Peter from Durham Johnston school, have made good starts on considering what happens for general $a$.

Both Peter and Pablo have made the good observatioin that $x=a$ will always be one of the points of intersection, because $x=a$ is always a solution to $x^a=a^x$.

But is it the case, for any value of $a$, that there are always two intersections of the graphs at positive $x$ values?

Sergio has solved completely the case where $0 < a \le 1$, and found there is only one intersection in this case.

In the case $0 < a \le 1$, If $x$ is smaller than $a$ then $a^x$ has a higher value than $x^a$, whereas if $x$ is bigger than $a$, $x^a$ would alway be bigger than $a^x$.

 

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Thomas from BHASVIC Sixth Form College Brighton has finished the solution for $a>1$:

At $x=0$ for $a > 1$ we have $0=x^a < a^x=1$. Think of the two graphs $y_1 = x^a$ and $y_2 = a^x$. As $x$ increases, we certainly have an intersection at $x=a$; the gradients of the curves here are $\frac{dy_1}{dx} = a^a$ and $\frac{dy_2}{dx} = a^a \ln a$ respectively. There are now 3 cases:

For $1 < a < e$ the gradient of $y_1$ is steeper than $y_2$ here, so $y_2$ must pass from above $y_1$ just before $x=a$ to below it just after $x=a$. We know that for large $x$, $y_2 > y_1$ so there must be a second intersection point somewhere for $x > a$.

For $a > e$ the gradient of $y_2$ is steeper than $y_1$ here, so $y_2$ passes from below $y_1$ before $x=a$ to above it after $x=a$. At $x=0$, $y_2 > y_1$ so the two curves must cross again somewhere in $0 < x < a$.

For $a=e$ the two curves just touch. There are no other intersections other than $x=a$.

So there is only one number $a > 1$ where we don't have at least two points where $a^x=x^a$.