Lösung 3.4:2a
Aus Online Mathematik Brückenkurs 1
(Unterschied zwischen Versionen)
			  			                                                      
		          
			| K  | K  (Robot: Automated text replacement  (-{{Displayed math +{{Abgesetzte Formel)) | ||
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| 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, | 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, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\ln 2^{x^2-2} = \ln 1\,,</math>}} | 
| 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, | 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, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\bigl(x^2-2\bigr)\ln 2 = \ln 1\,\textrm{.}</math>}} | 
| Because <math>e^{0}=1</math>, so <math>\ln 1 = 0</math>, giving | Because <math>e^{0}=1</math>, so <math>\ln 1 = 0</math>, giving | ||
| - | {{ | + | {{Abgesetzte Formel||<math>(x^2-2)\ln 2=0\,\textrm{.}</math>}} | 
| This means that ''x'' must satisfy the second-degree equation | This means that ''x'' must satisfy the second-degree equation | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x^2-2 = 0\,\textrm{.}</math>}} | 
| Taking the root gives <math>x=-\sqrt{2}</math> or <math>x=\sqrt{2}\,</math>. | Taking the root gives <math>x=-\sqrt{2}</math> or <math>x=\sqrt{2}\,</math>. | ||
Version vom 08:46, 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.
 
		  