Solution 3.4:2a

From Förberedande kurs i matematik 1

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

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.

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_3.4:2a"