## FANDOM

49 Pages

From NRICH,

find all real solutions to

$\left(2-x^2\right)^{x^2-3 \sqrt{2} x+4}=1\,$

Extension: What if x is permitted to be a complex number?

MathHelp replies,

### First tryEdit

ab=1 when a=1 or b=0 (and a is nonzero), so the first try at a solution is to solve

$x^2-3 \sqrt{2} x+4=0\,$

and

$2-x^2=1\,$

and take the union of the results. The first equation gives us $\sqrt{2}$ and $\sqrt{8}$. The second equation gives us 1 and −1. But then we need to throw out $\sqrt{2}$ because that solution gives us 00, which is undefined. So the results of the first try are:

$[itex]\sqrt{8}, \ 1, \ \mbox{and} \ 1. \,$

### Second tryEdit

Noting that (-1)2=1 and that $e^{2 i \pi k}=1\,$ for all integers, k, there are certainly other solutions to ab=1. For all I know at this point, there may be real solutions in x to the pair of equations

$x^2-3 \sqrt{2} x+4=a\,$

and

$2-x^2=b,\,$

where a and b are complex solutions to ab=1.

So I would like to start by characterizing solutions to ab=1. First, the equation is solved whenever a=1, regardless of b. If a is not equal to 1, then,

ab = eb ln(a) = $\mathrm{e}^{b \ln{a} - 2 i \pi k}=1\,$

and so the complete solution to ab=1 is

$a=1,\,$
$a \ne 0, \ a \ne 1, \ b = \frac{2 i \pi k}{\ln(a)}\,$

The second solution can also be expressed as

$a = e^{\frac{2 i \pi k}{b} }\,$