Ling Xiang Ning of Raffles Institution in
Singapore, found two solutions, and after trying some more values
of x, decided that there would be no more.
Vassil Vassilev, who is in year 11 at
Lawnswood High School in Leeds, used a graph to convince himself
that there were no more solutions.
First of all, he rearranged the equation:
#set var="tmp" value="$TeX"
#set var="TeX" value="l2h"
#set var="TeX" value="$tmp
$$ \eqalign {&x &=& (\sqrt2)^x & \\
&\left(\sqrt[x]{x}\right) &=& \sqrt2 } $$ (taking the
xth root)\\ $$ \eqalign & \left(\sqrt[x]{x}\right)^2
&=& 2(squaring) \\ &x^{\frac{2}{x}} &=& 2
}$$
Next he calculated values of x (2/x) for values of x
from 1 to 8, and plotted this graph:
Vassil commented that by looking at the original equation we
could rule out negative values of x, and that the values in the
graph decline after x=4. The justification for the decline is that
we are calculating smaller and smaller powers. However, you may not
be convinced, as the number we are finding powers of is
getting bigger. Are you sure that curve isn't going to go up again
further along?
The use of a graph to justify there being only 2 solutions was a
good idea.
Clearly $x=0$ is not a solution and there are no negative
solutions because the right hand side is positive. We have found 2
solutions $x=2$ and $x=4$ and we have to show that there are no
other solutions.
Consider the function $f(x) = (\sqrt 2)^x = exp(x\ln \sqrt
2)$. Differentiating the function gives: $$\frac{df}{dx} = (ln
\sqrt 2)\times exp(x\ln \sqrt 2)$$ and $$\frac{d^2f}{dx^2} = (ln
\sqrt 2)^2\times exp(x\ln \sqrt 2)> 0.$$ So the graph of f is
convex (the first derivative, or gradient, is always increasing) so
the graph of $y=f(x)$ meets $y=x$ in at most 2 points. So 2 and 4
are the only solutions.