Question

How do you evaluate the given logarithmic expression: ${{\log }_{9}}243$?

Answer 1

Hint: We start solving the problem by equating the given expression to a variable. We then make use of the results $9={{3}^{2}}$ and $243={{3}^{5}}$ to proceed through the problem. We then make use of the result ${{\log }_{{{a}^{m}}}}{{b}^{n}}=\dfrac{n}{m}{{\log }_{a}}b$ to proceed further through the problem. We then make use of the result ${{\log }_{a}}a=1$ and make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the value of the given logarithmic expression: ${{\log }_{9}}243$.
Let us assume $s={{\log }_{9}}243$ ---(1).
We know that $9={{3}^{2}}$ and $243={{3}^{5}}$. Let us use these results in equation (1).
$\Rightarrow s={{\log }_{{{3}^{2}}}}{{3}^{5}}$ ---(2).
We know that ${{\log }_{{{a}^{m}}}}{{b}^{n}}=\dfrac{n}{m}{{\log }_{a}}b$. Let us use this result in equation (2).
$\Rightarrow s=\dfrac{5}{2}\times {{\log }_{3}}3$ ---(3).
We know that ${{\log }_{a}}a=1$. Let us use this result in equation (3).
$\Rightarrow s=\dfrac{5}{2}\times 1$.
$\Rightarrow s=\dfrac{5}{2}$.
So, we have found the value of the logarithmic expression ${{\log }_{9}}243$ as $\dfrac{5}{2}$.
$\therefore $ The value of the given logarithmic expression ${{\log }_{9}}243$ is $\dfrac{5}{2}$.

Note:
We can also solve the given problem as shown below:
Now, we have $s={{\log }_{9}}243$.
$\Rightarrow s={{\log }_{9}}\left( 81\times 3 \right)$---(4).
We know that ${{\log }_{c}}\left( a\times b \right)={{\log }_{c}}a+{{\log }_{c}}b$. Let us use this result in equation (4).
$\Rightarrow s={{\log }_{9}}81+{{\log }_{9}}3$ ---(5).
We know that \[81={{9}^{2}}\] and $3={{9}^{\dfrac{1}{2}}}$. Let us use these results in equation (5).
$\Rightarrow s={{\log }_{9}}{{9}^{2}}+{{\log }_{9}}{{9}^{\dfrac{1}{2}}}$ ---(6).
We know that ${{\log }_{b}}{{a}^{n}}=n{{\log }_{b}}a$. Let us use this result in equation (6).
$\Rightarrow s=2{{\log }_{9}}9+\dfrac{1}{2}{{\log }_{9}}9$ ---(7).
We know that ${{\log }_{a}}a=1$. Let us use this result in equation (7).
$\Rightarrow s=2+\dfrac{1}{2}$.
$\Rightarrow s=\dfrac{4+1}{2}$.
$\Rightarrow s=\dfrac{5}{2}$.
So, the value of the logarithmic expression ${{\log }_{9}}243$ as $\dfrac{5}{2}$.