From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	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,
-	<center> [[Bild:3_4_2a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\ln 2^{x^2-2} = \ln 1\,,</math>}}
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		+	and use the log law <math>\ln a^b = b\cdot \ln a</math> to get the exponent <math>x^2-2</math> as a factor on the left-hand side,
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		+	{{Displayed math||<math>\bigl(x^2-2\bigr)\ln 2 = \ln 1\,\textrm{.}</math>}}
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		+	Because <math>e^{0}=1</math>, so <math>\ln 1 = 0</math>, giving
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		+	{{Displayed math||<math>(x^2-2)\ln 2=0\,\textrm{.}</math>}}
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		+	This means that ''x'' must satisfy the second-degree equation
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		+	{{Displayed math||<math>x^2-2 = 0\,\textrm{.}</math>}}
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		+	Taking the root gives <math>x=-\sqrt{2}</math> or <math>x=\sqrt{2}\,</math>.
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		+	Note: The exercise is taken from a Finnish upper-secondary final examination from March 2007.

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Current revision

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.