STEP I 1997 question 8 solution

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Clearly, the equation has no non-positive roots (as the RHS is always positive for real x).

Then we can take logs of both sides for the equivalent

.

Let . Then

Note that is strictly decreasing (well for positive x anyway, which are the only ones we're worrying about).

Now, setting we get

This is a maximum since was decreasing. The value at the maximum is

For there to be zeroes to f(x), we must have that the maximum is larger than or equal to zero:

Exponentiating:

and again:

,

and so there are no real roots if .

If , then and so is strictly positive (again for x>0). This means that there can be at most 1 solution to the equation. Consider now

which is negative since and . Then consider f(1):

which is positive since . Because f(x) is continuos, there must then be one point x, with such that , i.e. the equation has at least one solution.

But now must have exactly one root. Since this equation is equivalent (for positive x) to the given equation , it follows that this latter one also has exactly one solution.

Solution by ukgea