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# How Many Solutions?

This is a 'quickie'.

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 because:

"The graphs of $y = x$ and $y = (\sqrt {2})^x$ cut at exactly two points $(2, 2)$ and $(4, 4)$ so there are exactly two solutions."

Next he calculated values of $x^{\frac{2}{x}}$ from $1$ to $8$, and plotted this graph:

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This is a 'quickie'.

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 because:

"The graphs of $y = x$ and $y = (\sqrt {2})^x$ cut at exactly two points $(2, 2)$ and $(4, 4)$ so there are exactly two solutions."

Vassil from Lawnswood High School in
Leeds, used a different graph to convince himself that there were
no more solutions.

First of all, he rearranged the equation:

$$ \eqalign { x &=&
(\sqrt{2})^x \\ \sqrt[x]{x} &=& \sqrt{2} }$$ (taking the
$x^{th}$ root)

$$ \eqalign { (\sqrt[x]{x})^2
&=& 2 \\ x^{\frac{2}{x}} &=& 2 } $$
(squaring).

Next he calculated values of $x^{\frac{2}{x}}$ 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.