Aus Online Mathematik Brückenkurs 1

The left-hand side is " \displaystyle \text{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 \lg a^{b}=b\centerdot \lg a to get the exponent \displaystyle x^{\text{2}}-\text{2 } as a factor on the left-hand side

\displaystyle \left( x^{\text{2}}-\text{2 } \right)\ln 2=\ln 1

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

\displaystyle \left( x^{\text{2}}-\text{2 } \right)\ln 2=0

This means that \displaystyle x must satisfy the second-degree equation

\displaystyle \left( x^{\text{2}}-\text{2 } \right)=0

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.