Lösung 3.3:6c

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
Before we even start thinking about transforming
+
Before we even start thinking about transforming <math>\log_2</math> and <math>\log_3</math> to ln, we use the log laws
-
<math>\log _{2}</math>
+
-
and
+
-
<math>\log _{3}</math>
+
-
to ln, we use the log laws
+
-
 
+
-
 
+
-
<math>\lg a^{b}=b\centerdot \lg a</math>
+
-
 
+
-
 
+
-
<math>\lg \left( a\centerdot b \right)=\lg a+\lg b</math>
+
 +
{{Displayed math||<math>\begin{align}
 +
\log a^b &= b\cdot\log a\,,\\[5pt]
 +
\log (a\cdot b) &= \log a+\log b\,,
 +
\end{align}</math>}}
to simplify the expression
to simplify the expression
 +
{{Displayed math||<math>\begin{align}
 +
\log_{3}\log _{2}3^{118}
 +
&= \log_{3}(118\cdot\log_{2}3)\\[5pt]
 +
&= \log_{3}118 + \log_{3}\log_{2}3\,\textrm{.}
 +
\end{align}</math>}}
-
<math>\begin{align}
+
With help of the relation <math>2^{\log_{2}x} = x</math> and <math>3^{\log_{3}x} = x</math> and taking the natural logarithm , we can express <math>\log_{2}</math> and <math>\log_{3}</math> using ln,
-
& \log _{3}\log _{2}3^{118}=\log _{3}\left( 118\centerdot \log _{2}3 \right) \\
+
-
& =\log _{3}118+\log _{3}\log _{2}3 \\
+
-
\end{align}</math>
+
 +
{{Displayed math||<math>\log_{2}x=\frac{\ln x}{\ln 2}\quad</math> and <math>\quad\log_{3}x = \frac{\ln x}{\ln 3}\,\textrm{.}</math>}}
-
With help of the relation
+
The two terms <math>\log_3 118</math> and <math>\log_3\log_2 3</math> can therefore be written as
 +
{{Displayed math||<math>\log_{3}118 = \frac{\ln 118}{\ln 3}\quad</math> and <math>\quad\log_{3}\log_{2}3 = \log_{3}\frac{\ln 3}{\ln 2}\,,</math>}}
-
<math>2^{\log _{2}x}=x</math>
+
where we can simplify the last expression further with the logarithm law, log (a/b) = log a – log b, and then transform <math>\log _{3}</math> to ln,
-
and
+
-
<math>3^{\log _{3}x}=x</math>
+
-
+
-
 
+
-
and taking the natural logarithm , we can express
+
-
<math>\log _{2}</math>
+
-
and
+
-
<math>\log _{3}</math>
+
-
using ln,
+
-
 
+
-
 
+
-
<math>\log _{2}x=\frac{\ln x}{\ln 2}</math>
+
-
and
+
-
<math>\log _{3}x=\frac{\ln x}{\ln 3}</math>
+
-
 
+
-
 
+
-
 
+
-
The two terms log3 118 and log3 log2 3 can therefore be written as
+
-
 
+
-
 
+
-
<math>\log _{3}118=\frac{\ln 118}{\ln 3}</math>
+
-
and
+
-
<math>\log _{3}\log _{2}3=\log _{3}\left( \frac{\ln 3}{\ln 2} \right),</math>
+
-
 
+
-
 
+
-
where we can simplify the last expression further with the logarithm law, log a/b = log a – log b, and then transform
+
-
<math>\log _{3}</math>
+
-
to ln,
+
-
 
+
-
 
+
-
<math>\begin{align}
+
-
& \log _{3}\left( \frac{\ln 3}{\ln 2} \right)=\log _{3}\ln 3-\log _{3}\ln 2 \\
+
-
& =\frac{\ln \ln 3}{\ln 3}-\frac{\ln 2}{\ln 3} \\
+
-
\end{align}</math>
+
 +
{{Displayed math||<math>\begin{align}
 +
\log_{3}\frac{\ln 3}{\ln 2}
 +
&= \log_{3}\ln 3 - \log_{3}\ln 2\\[5pt]
 +
&= \frac{\ln\ln 3}{\ln 3} - \frac{\ln\ln 2}{\ln 3}\,\textrm{.}
 +
\end{align}</math>}}
In all, we thus obtain
In all, we thus obtain
-
 
+
{{Displayed math||<math>\log_{3}\log_{2}3^{118} = \frac{\ln 118}{\ln 3} + \frac{\ln \ln 3}{\ln 3} - \frac{\ln\ln 2}{\ln 3}\,\textrm{.}</math>}}
-
<math>\log _{3}\log _{2}3^{118}=\frac{\ln 118}{\ln 3}+\frac{\ln \ln 3}{\ln 3}-\frac{\ln 2}{\ln 3}</math>
+
-
 
+
Input into the calculator gives
Input into the calculator gives
 +
{{Displayed math||<math>\log_{3}\log_{2}3^{118}\approx 4\textrm{.}762\,\textrm{.}</math>}}
-
<math>\log _{3}\log _{2}3^{118}\approx 4.762</math>
 
- 
- 
- 
-
NOTE: the button sequence on a calculator will be:
 
 +
Note: The button sequence on the calculator will be:
-
<math>\begin{align}
+
<center>
-
& \left[ 1 \right]\quad \left[ 1 \right]\quad \left[ 8 \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ + \right] \\
+
{|
-
& \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ - \right]\quad \left[ 2 \right] \\
+
||
-
& \left[ \text{LN} \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 3 \right]\quad \left[ \text{LN} \right]\quad \left[ = \right] \\
+
{| border="1" cellpadding="3" cellspacing="0"
-
& \quad \\
+
|width="30px" align="center"|1
-
\end{align}</math>
+
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|1
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|8
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|LN
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|÷
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|3
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|LN
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|+
 +
|}
 +
|-
 +
|height="7px"|
 +
|-
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|3
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|LN
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|LN
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|÷
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|3
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|LN
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|-
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|2
 +
|}
 +
|-
 +
|height="7px"|
 +
|-
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|LN
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|LN
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|÷
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|3
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|LN
 +
|}
 +
||&nbsp;&nbsp;
 +
||
 +
{| border="1" cellpadding="3" cellspacing="0"
 +
|width="30px" align="center"|=
 +
|}
 +
|}
 +
</center>

Version vom 08:40, 2. Okt. 2008

Before we even start thinking about transforming \displaystyle \log_2 and \displaystyle \log_3 to ln, we use the log laws

Vorlage:Displayed math

to simplify the expression

Vorlage:Displayed math

With help of the relation \displaystyle 2^{\log_{2}x} = x and \displaystyle 3^{\log_{3}x} = x and taking the natural logarithm , we can express \displaystyle \log_{2} and \displaystyle \log_{3} using ln,

Vorlage:Displayed math

The two terms \displaystyle \log_3 118 and \displaystyle \log_3\log_2 3 can therefore be written as

Vorlage:Displayed math

where we can simplify the last expression further with the logarithm law, log (a/b) = log a – log b, and then transform \displaystyle \log _{3} to ln,

Vorlage:Displayed math

In all, we thus obtain

Vorlage:Displayed math

Input into the calculator gives

Vorlage:Displayed math


Note: The button sequence on the calculator will be:


1
  
1
  
8
  
LN
  
÷
  
3
  
LN
  
+
3
  
LN
  
LN
  
÷
  
3
  
LN
  
-
  
2
LN
  
LN
  
÷
  
3
  
LN
  
=
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.3:6c“