Question

How do you evaluate ${{\log }_{3}}\left( 243 \right)?$

Answer 1

We have to evaluate ${{\log }_{3}}\left( 243 \right),$ so to do so we will learn the properties of the log. We will learn that log (xy) is written as log (x) + log (y). We will use $\log \left( \dfrac{x}{y} \right)=\log \left( x \right)-\log \left( y \right).$ And to simplify this, we will also use $\log \left( {{x}^{y}} \right)=y\log \left( x \right).$ We need to know that the value of log at its base is always 1. We will learn that exponential and log are both connected functions. We will use the above properties to find our problem.

Complete answer:
We are asked to evaluate ${{\log }_{3}}\left( 243 \right).$ To solve the problem, we will learn about how the log function behaves. We will first understand the log function and learn its various properties. Before we start the log, we should know that log is just the inverse function of the exponential function. It behaves most similar to that exponential function. Thus we know that when we multiply two terms with the same base then their power gets added up.
${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$
Similarly, when we multiply two terms in log function then the value of the log of those terms added up, that is,
$\log \left( xy \right)=\log \left( x \right)+\log \left( y \right).......\left( i \right)$
As we know that when we divide two terms with the same base then their power gets subtracted, that is
$\dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}$
Similarly, when we apply log on $\dfrac{x}{y}$ then their value gets subtracted. That is,
$\log \left( \dfrac{x}{y} \right)=\log x-\log y$
We also know that
${{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}$
So, similarly in the log,
$\log \left( {{x}^{y}} \right)=y\log x.......\left( ii \right)$
We also know that,
$\log \left( 1 \right)=0$
So, we will use these properties to evaluate our given problem. Now we have ${{\log }_{3}}\left( 243 \right).$
Now, we will first factor 243 into prime factors. So, we get,
$243=3\times 3\times 3\times 3\times 3$
So, it can be written as
$\Rightarrow 243={{3}^{5}}$
Now, using $243={{3}^{5}}$ we get,
${{\log }_{3}}\left( 243 \right)={{\log }_{3}}{{3}^{5}}$
Now using $\log {{x}^{y}}=y\log x,$ we get,
$\Rightarrow {{\log }_{3}}\left( 243 \right)=5{{\log }_{3}}\left( 3 \right)\left[ \text{As }{{\log }_{3}}\left( {{3}^{5}} \right)=5{{\log }_{3}}\left( 3 \right) \right]$
Now, we use the value of log at its base as 1, that is ${{\log }_{a}}a=1.$ So,
${{\log }_{3}}\left( 3 \right)=1$
Using this above, we get,
$\Rightarrow 5\times 1\left[ \text{As }{{\log }_{3}}\left( 3 \right)=1 \right]$
$\Rightarrow 5$
So, we get ${{\log }_{3}}\left( 243 \right)=5.$

Note: Remember to solve such a problem by following the properties. Some of the properties like $\log \left( xy \right)=\log \left( x \right)+\log \left( y \right)$ and $\log \left( \dfrac{x}{y} \right)=\log \left( x \right)-\log \left( y \right).$ Or else we will reach to a wrong answer. We will simplify step by step so that it is clear and we don’t get any errors in this way.