Solution 3.3:6c
From Förberedande kurs i matematik 1
m (Lösning 3.3:6c moved to Solution 3.3:6c: Robot: moved page) |
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- | {{ | + | Before we even start thinking about transforming |
- | < | + | <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> | ||
+ | |||
+ | |||
+ | to simplify the expression | ||
+ | |||
+ | |||
+ | <math>\begin{align} | ||
+ | & \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> | ||
+ | |||
+ | |||
+ | 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, | ||
+ | |||
+ | |||
+ | <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> | ||
+ | |||
+ | |||
+ | In all, we thus obtain | ||
+ | |||
+ | |||
+ | <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 | ||
+ | |||
+ | |||
+ | <math>\log _{3}\log _{2}3^{118}\approx 4.762</math> | ||
+ | |||
+ | |||
+ | |||
+ | NOTE: the button sequence on a calculator will be: | ||
+ | |||
+ | |||
+ | |||
+ | <math>\begin{align} | ||
+ | & \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] \\ | ||
+ | & \quad \\ | ||
+ | \end{align}</math> |
Revision as of 09:49, 26 September 2008
Before we even start thinking about transforming
lga
a
b
=lga+lgb
to simplify the expression
118
log23
=log3118+log3log23
With help of the relation
and taking the natural logarithm , we can express
The two terms log3 118 and log3 log2 3 can therefore be written as
ln2ln3
where we can simplify the last expression further with the logarithm law, log a/b = log a – log b, and then transform
ln2ln3
=log3ln3−log3ln2=ln3lnln3−ln3ln2
In all, we thus obtain
Input into the calculator gives
4
762
NOTE: the button sequence on a calculator will be:
1
1
8
LN
3
LN
+
3
LN
LN
3
LN
−
2
LN
LN
3
LN
=