Question

Find the value of log5 using the series expansion method.

Answer 1

Hint: We should note that in the question, that log5 is given and not ln5. Therefore, we should try to convert log to ln and then use the series expansion of ln(1+x) to obtain the answer to the given question.

Complete step-by-step answer:
We know that whenever the base of the logarithm is not specified we should assume the base to be 10. Thus, we have to find ${{\log }_{10}}5$. Now, we also know that for logarithms,
${{\log }_{b}}a=\dfrac{{{\log }_{c}}a}{{{\log }_{b}}c}..................(1.1)$
Therefore, taking a=5, b=10 and c=e in equation 1.1, we obtain
 $\log 5={{\log }_{10}}5=\dfrac{{{\log }_{e}}5}{{{\log }_{e}}10}=\dfrac{\ln 5}{\ln 10}..................(1.2)$
We know that the value of\[ln10\approx 2.3025.........(1.3)\].
We know that the series expansion of ln(x) for $x > \dfrac{1}{2}$ is given by
$\ln (x)=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( \dfrac{x-1}{x} \right)}^{n}}}{n}}=\dfrac{x-1}{x}+\dfrac{1}{2}{{\left( \dfrac{x-1}{x} \right)}^{2}}+...\text{ }.........\text{(1}\text{.4)}$
Thus, using x=5 in equation (1.3), we obtain
$\begin{align}
  & \ln (5)=\dfrac{5-1}{5}+\dfrac{1}{2}{{\left( \dfrac{5-1}{5} \right)}^{2}}+...\text{ } \\
 & \text{=}\dfrac{4}{5}\text{+}\dfrac{1}{2}{{\left( \dfrac{4}{5} \right)}^{2}}\text{+}\dfrac{1}{3}{{\left( \dfrac{4}{5} \right)}^{3}}\text{+}...\text{ } \\
 & \text{=0}\text{.8+0}\text{.5}\times {{\left( 0.8 \right)}^{2}}\text{+0}\text{.333}\times {{\left( 0.8 \right)}^{3}}+0.25\times {{\left( 0.8 \right)}^{4}}+0.2\times {{\left( 0.8 \right)}^{5}}... \\
 & \approx \text{0}\text{.8+0}\text{.32+0}\text{.1707+0}\text{.1024+0}\text{.0655=1}\text{.4586}.........\text{(1}\text{.5)} \\
\end{align}$
Therefore, using the values of ln5 and ln10 from equations (1.3) and (1.5) in equation (1.2), we obtain
$\log 5=\dfrac{\ln 5}{\ln 10}\approx \dfrac{1.4586}{2.3025}=0.6335$
Which is the required answer to the given question.

Note: We should note that the actual value of log5 is close to 0.6989. However, as we have used values only up to fifth power of 0.8 in the expansion in (1.5), we got a value similar to the correct answer but 0.6335 instead of 0.6989. However, by taking values of higher powers of 0.8 in equation (1.5), the value would get closer and closer to the correct value. Also, we should note that in equation (1.4), the formula of log expansion for $x > \dfrac{1}{2}$ should be used as the expansion formula has a different form for $x < \dfrac{1}{2}$.