STEP I 1997 question 8 solution

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STEP I 1997 question 8 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1997 question 8 solution

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

$\ln x = x\ln a$

$\ln x - x\ln a = 0$ .

Let $f(x) = \ln x - x\ln a$ . Then

$\displaystyle f'(x) = \frac{1}{x} - \ln a$

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

Now, setting f'(x) = 0 we get

$\displaystyle \frac{1}{x} = \ln a$

$\displaystyle x = \frac{1}{\ln a}$

This is a maximum since f'(x) was decreasing. The value at the maximum is

$\displaystyle f\left(\frac{1}{\ln a}\right) = - \ln {(\ln a)} - 1$

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

$- \ln{(\ln a)} - 1 \geq 0$

$\ln{(\ln a)} \leq -1$

Exponentiating:

$\ln a \leq e^{-1}$

and again:

$a \leq e^{1/e}$ ,

and so there are no real roots if $a > e^{1/e}$ .

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

$f(a) = \ln a - a\ln a = (1 - a)\ln a$

which is negative since 1 - a > 0 and $\ln a < 0$ . Then consider f(1):

$f(1) = \ln 1 - \ln a = - \ln a$

which is positive since $\ln a < 0$ . Because f(x) is continuos, there must then be one point x, with a < x < 1 such that f(x) = 0 , i.e. the equation has at least one solution.

But now f(x) = 0 must have exactly one root. Since this equation is equivalent (for positive x) to the given equation x = a^x , it follows that this latter one also has exactly one solution.

Solution by ukgea

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