Solution 3.4:3a

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 3.4:3a moved to Solution 3.4:3a: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
Both left- and right-hand sides are positive for all values of
-
<center> [[Image:3_4_3a-1(2).gif]] </center>
+
<math>x</math>
-
{{NAVCONTENT_STOP}}
+
and this means that we can take the logarithm of both sides and get a more manageable equation.
-
{{NAVCONTENT_START}}
+
 
-
<center> [[Image:3_4_3a-2(2).gif]] </center>
+
LHS
-
{{NAVCONTENT_STOP}}
+
<math>=\ln 2^{-x^{2}}=-x^{2}\centerdot \ln 2,</math>
 +
 
 +
RHS
 +
<math>=\ln \left( 2e^{2x} \right)=\ln 2+\ln e^{2x}=\ln 2+2x\centerdot \ln e=\ln 2+2x\centerdot 1</math>
 +
 
 +
 
 +
After a little rearranging, the equation becomes
 +
 
 +
 
 +
<math>x^{2}+\frac{2}{\ln 2}x+1=0</math>
 +
 
 +
We complete the square of the left-hand side
 +
 
 +
 
 +
<math>\left( x+\frac{1}{\ln 2} \right)^{2}-\left( \frac{1}{\ln 2} \right)^{2}+1=0</math>
 +
 
 +
 
 +
and move the constant terms over to the right-hand side:
 +
 
 +
 
 +
<math>\left( x+\frac{1}{\ln 2} \right)^{2}=\left( \frac{1}{\ln 2} \right)^{2}-1</math>
 +
 
 +
 
 +
It can be difficult to see whether the right-hand side is positive or not, but if we remember that
 +
<math>e>2</math>
 +
and that thus
 +
<math>\text{ln 2}<\ln e=\text{1}</math>, we must have that
 +
<math>\left( {1}/{\ln 2}\; \right)^{2}>1</math>, i.e. the right-hand side is positive.
 +
 
 +
The equation therefore has the solutions
 +
 
 +
 
 +
<math>x=-\frac{1}{\ln 2}\pm \sqrt{\left( \frac{1}{\ln 2} \right)^{2}-1,}</math>
 +
 
 +
 
 +
which can also be written as
 +
 
 +
 
 +
<math>x=\frac{-1\pm \sqrt{1-\left( \ln 2 \right)^{2}}}{\ln 2}</math>

Revision as of 11:41, 26 September 2008

Both left- and right-hand sides are positive for all values of \displaystyle x and this means that we can take the logarithm of both sides and get a more manageable equation.

LHS \displaystyle =\ln 2^{-x^{2}}=-x^{2}\centerdot \ln 2,

RHS \displaystyle =\ln \left( 2e^{2x} \right)=\ln 2+\ln e^{2x}=\ln 2+2x\centerdot \ln e=\ln 2+2x\centerdot 1


After a little rearranging, the equation becomes


\displaystyle x^{2}+\frac{2}{\ln 2}x+1=0

We complete the square of the left-hand side


\displaystyle \left( x+\frac{1}{\ln 2} \right)^{2}-\left( \frac{1}{\ln 2} \right)^{2}+1=0


and move the constant terms over to the right-hand side:


\displaystyle \left( x+\frac{1}{\ln 2} \right)^{2}=\left( \frac{1}{\ln 2} \right)^{2}-1


It can be difficult to see whether the right-hand side is positive or not, but if we remember that \displaystyle e>2 and that thus \displaystyle \text{ln 2}<\ln e=\text{1}, we must have that \displaystyle \left( {1}/{\ln 2}\; \right)^{2}>1, i.e. the right-hand side is positive.

The equation therefore has the solutions


\displaystyle x=-\frac{1}{\ln 2}\pm \sqrt{\left( \frac{1}{\ln 2} \right)^{2}-1,}


which can also be written as


\displaystyle x=\frac{-1\pm \sqrt{1-\left( \ln 2 \right)^{2}}}{\ln 2}

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_3.4:3a"