Question

Find the value of $\log 7$ using series expansion.

Answer 1

Hint: Use the fact that the expansion of function $y=\log x$ is $\log x=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( \dfrac{x-1}{x} \right)}^{n}}}{n}}$ for all $x>\dfrac{1}{2}$. Substitute $x=7$ in the above formula and simplify the expression to write the series expansion of $\log 7$.

Complete step-by-step solution -
We have to write the series expansion of $\log 7$.
We know that for all $x>\dfrac{1}{2}$, we can write the series expansion of $y=\log x$ as $\log x=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( \dfrac{x-1}{x} \right)}^{n}}}{n}}$.
Substituting $x=7$ in the above expression, we have $\log 7=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( \dfrac{7-1}{7} \right)}^{n}}}{n}}$.
Simplifying the above expression, we have $\log 7=\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( \dfrac{7-1}{7} \right)}^{n}}}{n}}=\dfrac{{{\left( \dfrac{6}{7} \right)}^{1}}}{1}+\dfrac{{{\left( \dfrac{6}{7} \right)}^{2}}}{2}+\dfrac{{{\left( \dfrac{6}{7} \right)}^{3}}}{3}+...=\dfrac{6}{7}+\dfrac{1}{2}{{\left( \dfrac{6}{7} \right)}^{2}}+\dfrac{1}{3}{{\left( \dfrac{6}{7} \right)}^{3}}+...$.
Hence, the series expansion of $\log 7$ is $\log 7=\dfrac{6}{7}+\dfrac{1}{2}{{\left( \dfrac{6}{7} \right)}^{2}}+\dfrac{1}{3}{{\left( \dfrac{6}{7} \right)}^{3}}+...$.
Logarithmic functions are inverses of exponential functions. It means that the logarithm of a given number ‘x’ is the exponent to which another fixed number, the base ‘b’, must be raised to produce that number ‘x’. We can also say that logarithm counts the number of occurrences of the same factor in repeated multiplication.
We can write a series expansion of any differentiable function using Taylor Series Expansion. Taylor series of a function is an infinite sum of terms that are expressed in terms of the function’s derivative at a single point. Taylor series expansion of any function $f\left( x \right)$ at the point $x=a$ is $f\left( a \right)+\dfrac{f'\left( a \right)}{1!}\left( x-a \right)+\dfrac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+...=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}}$.

Note: While writing the power series expansion of $y=\log x$, one must be careful about the domain of the function. Expansion of $y=\log x$ shows different behaviour for different values of ‘x’. If we calculate the expansion using one formula, we will get an incorrect answer.