Lösung 3.4:2a
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Version vom 14:41, 22. Okt. 2008
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.
