
How do you evaluate \[\ln 1.4\]?
Answer
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Hint: In this problem, we have to evaluate the natural log \[\ln 1.4\]. We can evaluate using two ways, the one is using calculators and the other is using Taylor series. In this problem we have to evaluate manually, so we are going to evaluate using Taylor series. We can split the given natural log value to apply in the Taylor series to get the natural log value.
Complete step by step answer:
We know that the given natural log to be evaluated is \[\ln 1.4\].
Now we can evaluate the given natural log using Taylor series.
We know that the Taylor series is,
\[\ln \left( 1+x \right)=\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n-1}}\dfrac{{{x}^{n}}}{n}}\] .
Now we can split 1.4 as 1 + 0.4, we get
\[\ln \left( 1.4 \right)=\ln \left( 1+0.4 \right)\]
Now we can apply the Taylor series for the above natural log, we get
\[\begin{align}
& \Rightarrow \ln \left( 1+0.4 \right)=0.4-\dfrac{{{\left( 0.4 \right)}^{2}}}{2}+\dfrac{{{\left( 0.4 \right)}^{3}}}{3}-\dfrac{{{\left( 0.4 \right)}^{4}}}{4}+...... \\
& \\
\end{align}\]
Now we can multiply with the respective of the exponents and divide to its denominator and add the final answer in the above step to get an approximate value, we get
\[\begin{align}
& \Rightarrow \ln \left( 1+0.4 \right)=0.4-\dfrac{{{\left( 0.4 \right)}^{2}}}{2}+\dfrac{{{\left( 0.4 \right)}^{3}}}{3}-\dfrac{{{\left( 0.4 \right)}^{4}}}{4}+\dfrac{{{\left( 0.4 \right)}^{5}}}{5}-\dfrac{{{\left( 0.4 \right)}^{6}}}{6}.... \\
& \approx 0.4-0.08+0.02133333-0.0064 \\
& +0.002048-0.00068267+0.00023406 \\
& -0.00008192+0.00002913-0.00001049 \\
& +0.00000381-0.00000140+0.00000052 \\
& -0.00000019+0.00000007-0.00000003+0.00000001 \\
& =0.33647223 \\
\end{align}\]
Therefore, the approximate value of \[\ln 1.4\] is 0.33647223.
Note: Students make mistakes while calculating the decimal part in the Taylor series. We should have some basic knowledge of natural logs and properties or formulas that can be used in these types of problems. Students should concentrate in the decimal part while squaring and dividing in the Taylor series. To solve these types of problems we should understand the concepts and properties of natural logs.
Complete step by step answer:
We know that the given natural log to be evaluated is \[\ln 1.4\].
Now we can evaluate the given natural log using Taylor series.
We know that the Taylor series is,
\[\ln \left( 1+x \right)=\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n-1}}\dfrac{{{x}^{n}}}{n}}\] .
Now we can split 1.4 as 1 + 0.4, we get
\[\ln \left( 1.4 \right)=\ln \left( 1+0.4 \right)\]
Now we can apply the Taylor series for the above natural log, we get
\[\begin{align}
& \Rightarrow \ln \left( 1+0.4 \right)=0.4-\dfrac{{{\left( 0.4 \right)}^{2}}}{2}+\dfrac{{{\left( 0.4 \right)}^{3}}}{3}-\dfrac{{{\left( 0.4 \right)}^{4}}}{4}+...... \\
& \\
\end{align}\]
Now we can multiply with the respective of the exponents and divide to its denominator and add the final answer in the above step to get an approximate value, we get
\[\begin{align}
& \Rightarrow \ln \left( 1+0.4 \right)=0.4-\dfrac{{{\left( 0.4 \right)}^{2}}}{2}+\dfrac{{{\left( 0.4 \right)}^{3}}}{3}-\dfrac{{{\left( 0.4 \right)}^{4}}}{4}+\dfrac{{{\left( 0.4 \right)}^{5}}}{5}-\dfrac{{{\left( 0.4 \right)}^{6}}}{6}.... \\
& \approx 0.4-0.08+0.02133333-0.0064 \\
& +0.002048-0.00068267+0.00023406 \\
& -0.00008192+0.00002913-0.00001049 \\
& +0.00000381-0.00000140+0.00000052 \\
& -0.00000019+0.00000007-0.00000003+0.00000001 \\
& =0.33647223 \\
\end{align}\]
Therefore, the approximate value of \[\ln 1.4\] is 0.33647223.
Note: Students make mistakes while calculating the decimal part in the Taylor series. We should have some basic knowledge of natural logs and properties or formulas that can be used in these types of problems. Students should concentrate in the decimal part while squaring and dividing in the Taylor series. To solve these types of problems we should understand the concepts and properties of natural logs.
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