Solution 3.4:1b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (08:54, 2 October 2008) (edit) (undo)
m
 
Line 1: Line 1:
-
In the equation, both sides are positive because the factors
+
In the equation, both sides are positive because the factors <math>e^{x}</math> and <math>3^{-x}</math> are positive regardless of the value of <math>x</math> (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides,
-
<math>e^{x}</math>
+
-
and
+
-
<math>3^{-x}</math>
+
-
are positive regardless of the value of
+
-
<math>x</math>
+
-
(a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both numbers,
+
 +
{{Displayed math||<math>\ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}</math>}}
-
<math>\ln \left( 13e^{x} \right)=\ln \left( 2\centerdot 3^{-x} \right)</math>
+
Using the log laws, we can divide up the products into several logarithmic terms,
 +
{{Displayed math||<math>\ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},</math>}}
-
Using the log law, we can divide up the products into several logarithmic terms,
+
and using the law <math>\ln a^{b}=b\cdot \ln a</math>, we can get rid of <math>x</math> from the exponents
 +
{{Displayed math||<math>\ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}</math>}}
-
<math>\ln 13+\ln e^{x}=\ln 2+\ln 3^{-x}</math>
+
Collect <math>x</math> on one side and the other terms on the other,
 +
{{Displayed math||<math>x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}</math>}}
-
and using the law
+
Take out <math>x</math> on the left-hand side and use <math>\ln e=1</math>,
-
<math>\ln a^{b}=b\centerdot \ln a</math>, we can get rid of
+
-
<math>x</math>
+
-
from the exponents:
+
 +
{{Displayed math||<math>x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}</math>}}
-
<math>\ln 13+x\ln e=\ln 2+\left( -x \right)\ln 3</math>
+
Then, solve for <math>x</math>,
 +
{{Displayed math||<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}</math>}}
-
Collecting together
 
-
<math>x</math>
 
-
on one side and the other terms on the other,
 
 +
Note: Because <math>\ln 2 < \ln 13</math>, we can write the answer as
-
<math>x\ln e+x\ln 3=\ln 2-\ln 13</math>
+
{{Displayed math||<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>}}
-
 
+
to indicate that <math>x</math> is negative.
-
Take out
+
-
<math>x</math>
+
-
on the left-hand side and use
+
-
<math>\ln e=1</math>
+
-
:
+
-
 
+
-
 
+
-
<math>x\left( 1+\ln 3 \right)=\ln 2-\ln 13</math>
+
-
 
+
-
 
+
-
Then, solve for
+
-
<math>x</math>
+
-
:
+
-
 
+
-
 
+
-
<math>x=\frac{\ln 2-\ln 13}{1+\ln 3}</math>
+
-
 
+
-
 
+
-
NOTE: Because
+
-
<math>\ln 2<\ln 13</math>, we can write the answer as
+
-
 
+
-
 
+
-
<math>x=-\frac{\ln 13-\ln 2}{1+\ln 3}</math>
+
-
 
+
-
 
+
-
in order to indicate that
+
-
<math>x</math>
+
-
is negative.
+

Current revision

In the equation, both sides are positive because the factors \displaystyle e^{x} and \displaystyle 3^{-x} are positive regardless of the value of \displaystyle x (a positive base raised to a number always gives a positive number). We can therefore take the natural logarithm of both sides,

\displaystyle \ln\bigl(13e^{x}\bigr) = \ln\bigl(2\cdot 3^{-x}\bigr)\,\textrm{.}

Using the log laws, we can divide up the products into several logarithmic terms,

\displaystyle \ln 13+\ln e^{x} =\ln 2+\ln 3^{-x},

and using the law \displaystyle \ln a^{b}=b\cdot \ln a, we can get rid of \displaystyle x from the exponents

\displaystyle \ln 13 + x\ln e = \ln 2 + (-x)\ln 3\,\textrm{.}

Collect \displaystyle x on one side and the other terms on the other,

\displaystyle x\ln e+x\ln 3=\ln 2-\ln 13\,\textrm{.}

Take out \displaystyle x on the left-hand side and use \displaystyle \ln e=1,

\displaystyle x( 1+\ln 3)=\ln 2-\ln 13\,\textrm{.}

Then, solve for \displaystyle x,

\displaystyle x=\frac{\ln 2-\ln 13}{1+\ln 3}\,\textrm{.}


Note: Because \displaystyle \ln 2 < \ln 13, we can write the answer as

\displaystyle x=-\frac{\ln 13-\ln 2}{1+\ln 3}

to indicate that \displaystyle x is negative.

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_3.4:1b"