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Solution 3.4:2a

From Förberedande kurs i matematik 1

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Current revision (11:15, 2 October 2008) (edit) (undo)
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The left-hand side is "
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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,
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<math>\text{2}</math>
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raised to something", and therefore a positive number regardless of whatever value the exponent has. We can therefore take the log of both sides,
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{{Displayed math||<math>\ln 2^{x^2-2} = \ln 1\,,</math>}}
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<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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and use the log law
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{{Displayed math||<math>\bigl(x^2-2\bigr)\ln 2 = \ln 1\,\textrm{.}</math>}}
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<math>\lg a^{b}=b\centerdot \lg a</math>
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to get the exponent
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<math>x^{\text{2}}-\text{2 }</math>
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as a factor on the left-hand side
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Because <math>e^{0}=1</math>, so <math>\ln 1 = 0</math>, giving
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<math>\left( x^{\text{2}}-\text{2 } \right)\ln 2=\ln 1</math>
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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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Because
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{{Displayed math||<math>x^2-2 = 0\,\textrm{.}</math>}}
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<math>e^{0}=1</math>, so
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<math>\text{ln 1}=0</math>, giving:
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Taking the root gives <math>x=-\sqrt{2}</math> or <math>x=\sqrt{2}\,</math>.
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<math>\left( x^{\text{2}}-\text{2 } \right)\ln 2=0</math>
 
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Note: The exercise is taken from a Finnish upper-secondary final examination from March 2007.
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This means that
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<math>x</math>
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must satisfy the second-degree equation
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<math>\left( x^{\text{2}}-\text{2 } \right)=0</math>
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Taking the root gives
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<math>x=-\sqrt{2}</math>
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or
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<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,

ln2x2−2=ln1

and use the log law lnab=blna to get the exponent x2−2 as a factor on the left-hand side,

x2−2ln2=ln1. 

Because e0=1, so ln1=0, giving

(x2−2)ln2=0.

This means that x must satisfy the second-degree equation

x2−2=0.

Taking the root gives x=−2  or x=2 .


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

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