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# Find the value of $\log 3$ using series expansion.

Last updated date: 17th Jul 2024
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Hint: We know that the expansion of a function $y = \ln x$ is $\ln x = \sum\limits_{n = 1}^\infty {\dfrac{{{{\left( {\dfrac{{x - 1}}{x}} \right)}^n}}}{n}}$ for all $x > \dfrac{1}{2}$. Substitute $x = 3$ in the above given expansion and simplify the expression to write the value of $\ln 3$ using series expansion. But we have to calculate the value of $\log 3$. So, we should convert $\log 3$ into $\ln 3$. Then, applying this formula we get the required value of $\log 3$.

Here, we have to write the value of $\log 3$ by series expansion.
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 the value of ${\log _{10}}3$.
We also know that ${\log _b}a = \dfrac{{{{\log }_c}a}}{{{{\log }_b}c}} - - - - -$
Therefore, taking $a = 3$, $b = 10$ and $c = e$ in the above equation, we get
$\log 3 = {\log _{10}}3 = \dfrac{{{{\log }_e}3}}{{{{\log }_e}10}} = \dfrac{{\ln 3}}{{\ln 10}} - - - -$.
We know that the value of $\ln 10 \approx 2.3025$
We know that for all $x > \dfrac{1}{2}$, we can write the series expansion of $y = \ln x$ as $\ln x = \sum\limits_{n = 1}^\infty {\dfrac{{{{\left( {\dfrac{{x - 1}}{x}} \right)}^n}}}{n}}$.
Substituting $x = 3$ in the above expression, we have $\ln 3 = \sum\limits_{n = 1}^\infty {\dfrac{{{{\left( {\dfrac{{3 - 1}}{3}} \right)}^n}}}{n}}$.
Now, expanding and simplifying the above expression, we get
$\ln 3 = \sum\limits_{n = 1}^\infty {\dfrac{{{{\left( {\dfrac{{3 - 1}}{3}} \right)}^n}}}{n}} = \dfrac{{{{\left( {\dfrac{2}{3}} \right)}^1}}}{1} + \dfrac{{{{\left( {\dfrac{2}{3}} \right)}^2}}}{2} + \dfrac{{{{\left( {\dfrac{2}{3}} \right)}^3}}}{3} + \dfrac{{{{\left( {\dfrac{2}{3}} \right)}^4}}}{4} +$ - - - - - - - - - - - - - - - - - - - - - -
$\ln 3 = \dfrac{2}{3} + \dfrac{1}{2}{\left( {\dfrac{2}{3}} \right)^2} + \dfrac{1}{3}{\left( {\dfrac{2}{3}} \right)^3} + \dfrac{1}{4}{\left( {\dfrac{2}{3}} \right)^4} +$- - - - - - - - - - - - - - - - - -
So, the series expansion of $\ln 3$ is $\ln 3 = \dfrac{2}{3} + \dfrac{1}{2}{\left( {\dfrac{2}{3}} \right)^2} + \dfrac{1}{3}{\left( {\dfrac{2}{3}} \right)^3} + \dfrac{1}{4}{\left( {\dfrac{2}{3}} \right)^4} +$- - - - - -.
Now, we have to find the value of $\log 3$.
$\Rightarrow \ln 3 = 0.666 + 0.2217 + 0.0984 + 0.0491 + 0.0262 +$- - - - - - -
$\therefore \ln 3 = 1.0614$- - - - - -.
Now, $\log 3 = \dfrac{{\ln 3}}{{\ln 10}} = \dfrac{{1.0614}}{{2.3025}} \approx 0.47$.

Thus, the required value of $\log 3$ is $0.47$.

Note: While writing the power series expansion of $y = \ln x$, one must be careful about the domain of the function. Expansion of $y = \ln x$ shows different behaviour for different value of $x$.