From Förberedande kurs i matematik 1

The left-hand side is "2 raised to something", and therefore a positive number regardless of whatever value the exponent has. We can therefore take the log of both sides,

\displaystyle \ln 2^{x^2-2} = \ln 1\,,

and use the log law \displaystyle \ln a^b = b\cdot \ln a to get the exponent \displaystyle x^2-2 as a factor on the left-hand side,

\displaystyle \bigl(x^2-2\bigr)\ln 2 = \ln 1\,\textrm{.}

Because \displaystyle e^{0}=1, so \displaystyle \ln 1 = 0, giving

\displaystyle (x^2-2)\ln 2=0\,\textrm{.}

This means that x must satisfy the second-degree equation

\displaystyle x^2-2 = 0\,\textrm{.}

Taking the root gives \displaystyle x=-\sqrt{2} or \displaystyle x=\sqrt{2}\,.

Note: The exercise is taken from a Finnish upper-secondary final examination from March 2007.