Question

What will be the value of ${{\log }_{2}}\left( {{\log }_{3}}81 \right)$ ?

Answer 1

Hint: We will be using the concept of logarithmic functions to solve the problem. We will be using logarithmic identities like ${{\log }_{a}}{{b}^{m}}=m{{\log }_{a}}b$ to further simplify the problem also we will be using concepts of exponentiation to solve the problem.

Complete step-by-step solution -
Now, we have been given the expression ${{\log }_{2}}\left( {{\log }_{3}}81 \right)$and we have to find its value.
We will first simplify log 81 and then use its nature to find the nature of ${{\log }_{2}}\left( {{\log }_{3}}81 \right)$.
Now, we know that 81 can be represented as ${{3}^{4}}$ .
 Therefore, we can write ${{\log }_{3}}81={{\log }_{3}}{{3}^{4}}$ .
Now we know that ${{\log }_{a}}{{b}^{m}}=m{{\log }_{a}}b$. Therefore,
${{\log }_{3}}{{3}^{4}}=4{{\log }_{3}}3$.
Now we know that ${{\log }_{a}}a=1$ . Therefore,
${{\log }_{3}}{{3}^{4}}=4$ ……………….. (1)
Now we will use the value of ${{\log }_{3}}{{3}^{4}}$ from (1) in ${{\log }_{2}}\left( {{\log }_{3}}81 \right)$. So, we have,
${{\log }_{2}}\left( {{\log }_{3}}81 \right)={{\log }_{2}}4$
Also we can write 4 as ${{2}^{2}}$ . Therefore,
$={{\log }_{2}}{{2}^{2}}$ .
Again we will use the identity that ${{\log }_{a}}{{b}^{m}}=m{{\log }_{a}}b$. Therefore,
$=2{{\log }_{2}}2$ .
Also we know that ${{\log }_{2}}2=1$ . Therefore,
${{\log }_{2}}\left( {{\log }_{3}}81 \right)=2$.

Note: To solve these types of questions one must know some basic logarithmic identities like
$\begin{align}
  & {{\log }_{a}}{{b}^{m}}=m{{\log }_{a}}b \\
 & {{\log }_{a}}{{m}^{b}}=\dfrac{1}{m}{{\log }_{a}}^{b} \\
 & {{\log }_{a}}{a}=1 \\
 & {{\log }_{a}1}=0 \\
\end{align}$