Lösung 3.3:6b
Aus Online Mathematik Brückenkurs 1
K (Lösning 3.3:6b moved to Solution 3.3:6b: Robot: moved page) |
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- | {{ | + | The logarithm |
- | < | + | <math>\text{lg 46 }</math> |
- | {{ | + | satisfies the relation |
- | [[ | + | |
+ | |||
+ | <math>\text{10}^{\text{lg 46 }}=46</math> | ||
+ | |||
+ | |||
+ | and taking the natural logarithm of both sides, we obtain | ||
+ | |||
+ | |||
+ | <math>\ln \text{10}^{\text{lg 46 }}=\ln 46</math> | ||
+ | |||
+ | |||
+ | If we use the logarithm law, | ||
+ | <math>\lg a^{b}=b\centerdot \lg a</math>, on the left-hand side, the equality becomes | ||
+ | |||
+ | |||
+ | <math>\lg 46\centerdot \ln 10=\ln 46</math> | ||
+ | |||
+ | |||
+ | This shows that | ||
+ | |||
+ | |||
+ | <math>\lg 46=\frac{\ln 46}{\ln 10}=\frac{3.828641}{2.302585}=1.6627578</math> | ||
+ | |||
+ | |||
+ | and the answer is | ||
+ | <math>\text{1}.\text{663}</math>. | ||
+ | |||
+ | NOTE: In order to calculate the answer on a calculator, you press | ||
+ | |||
+ | |||
+ | <math>\begin{align} | ||
+ | & \left[ 4 \right]\quad \left[ 6 \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 1 \right]\quad \left[ 0 \right]\quad \left[ \text{LN} \right]\quad \left[ = \right] \\ | ||
+ | & \quad \\ | ||
+ | \end{align}</math> |
Version vom 09:22, 26. Sep. 2008
The logarithm \displaystyle \text{lg 46 } satisfies the relation
\displaystyle \text{10}^{\text{lg 46 }}=46
and taking the natural logarithm of both sides, we obtain
\displaystyle \ln \text{10}^{\text{lg 46 }}=\ln 46
If we use the logarithm law,
\displaystyle \lg a^{b}=b\centerdot \lg a, on the left-hand side, the equality becomes
\displaystyle \lg 46\centerdot \ln 10=\ln 46
This shows that
\displaystyle \lg 46=\frac{\ln 46}{\ln 10}=\frac{3.828641}{2.302585}=1.6627578
and the answer is
\displaystyle \text{1}.\text{663}.
NOTE: In order to calculate the answer on a calculator, you press
\displaystyle \begin{align}
& \left[ 4 \right]\quad \left[ 6 \right]\quad \left[ \text{LN} \right]\quad \left[ \div \right]\quad \left[ 1 \right]\quad \left[ 0 \right]\quad \left[ \text{LN} \right]\quad \left[ = \right] \\
& \quad \\
\end{align}