Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Solution 3.4:2b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 3.4:2b moved to Solution 3.4:2b: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
If we write the equation as
-
<center> [[Image:3_4_2b-1(2).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
-
{{NAVCONTENT_START}}
+
 
-
<center> [[Image:3_4_2b-2(2).gif]] </center>
+
<math>\left( e^{x} \right)^{2}+e^{x}=4</math>
-
{{NAVCONTENT_STOP}}
+
 
 +
 
 +
we see that
 +
<math>x</math>
 +
appears only in the combination
 +
<math>e^{x}</math>
 +
and it is therefore appropriate to treat
 +
<math>e^{x}</math>
 +
as a new unknown in the equation and then, when we have obtained the value of
 +
<math>e^{x}</math>, we can calculate the corresponding value of
 +
<math>x</math>
 +
by simply taking the logarithm.
 +
 
 +
For clarity, we set
 +
<math>t=e^{x}</math>, so that the equation can be written as
 +
 
 +
 
 +
<math>t^{2}+t=4</math>
 +
 
 +
 
 +
and we solve this second-degree equation by completing the square,
 +
 
 +
 
 +
<math>t^{2}+t=\left( t+\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}=\left( t+\frac{1}{2} \right)^{2}-\frac{1}{4}</math>
 +
 
 +
 
 +
which gives
 +
 
 +
 
 +
<math>\left( t+\frac{1}{2} \right)^{2}-\frac{1}{4}=4\quad \Leftrightarrow \quad t=-\frac{1}{2}\pm \frac{\sqrt{17}}{2}</math>
 +
 
 +
 
 +
These two roots give us two possible values for
 +
<math>e^{x}</math>,
 +
 
 +
 
 +
<math>e^{x}=-\frac{1}{2}-\frac{\sqrt{17}}{2}</math>
 +
or
 +
<math>e^{x}=-\frac{1}{2}+\frac{\sqrt{17}}{2}</math>
 +
 
 +
 
 +
In the first case, the right-hand side is negative and because "
 +
<math>e</math>
 +
raised to anything" can never be negative, there is no
 +
<math>x</math>
 +
that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because
 +
<math>\sqrt{17}>1</math>
 +
) and we can take the logarithm of both sides to obtain
 +
 
 +
 
 +
<math>x=\ln \left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)</math>
 +
 
 +
 
 +
NOTE: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting
 +
<math>t=\frac{\sqrt{17}}{2}-\frac{1}{2}</math>
 +
into the equation
 +
<math>t^{\text{2}}+t=\text{4}</math>,
 +
 
 +
LHS
 +
<math>=</math>
 +
 
 +
 +
<math>\begin{align}
 +
& =\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)^{2}+\left( \frac{\sqrt{17}}{2}-\frac{1}{2} \right)=\frac{17}{4}-2\centerdot \frac{1}{2}\centerdot \frac{\sqrt{17}}{2}+\frac{1}{4}+\frac{\sqrt{17}}{2}-\frac{1}{2} \\
 +
& =\frac{17}{4}+\frac{1}{4}-\frac{1}{2}=\frac{17+1-2}{4}=\frac{16}{4}=4= \\
 +
\end{align}</math>
 +
 
 +
 
 +
<math>=</math>
 +
RHS.

Revision as of 11:44, 6 October 2008

If we write the equation as


ex2+ex=4 


we see that x appears only in the combination ex and it is therefore appropriate to treat ex as a new unknown in the equation and then, when we have obtained the value of ex, we can calculate the corresponding value of x by simply taking the logarithm.

For clarity, we set t=ex, so that the equation can be written as


t2+t=4


and we solve this second-degree equation by completing the square,


t2+t=t+212−212=t+212−41 


which gives


t+212−41=4t=−21217 


These two roots give us two possible values for ex,


ex=−21−217  or ex=−21+217 


In the first case, the right-hand side is negative and because " e raised to anything" can never be negative, there is no x that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because 171  ) and we can take the logarithm of both sides to obtain


x=ln217−21 


NOTE: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting t=217−21  into the equation t2+t=4,

LHS =


=217−212+217−21=417−221217+41+217−21=417+41−21=417+1−2=416=4= 


= RHS.

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