Question

How do you evaluate ${{\log }_{a}}x=\dfrac{3}{2}{{\log }_{a}}9+{{\log }_{a}}2$?

Answer 1

Hint: In this problem we need to solve the given equation. We can observe that the given equation has logarithmic function. So, we are going to consider the RHS part of the given equation. For the value $\dfrac{3}{2}{{\log }_{a}}9$ we will use the logarithmic formula $m\log n=\log {{n}^{m}}$ and use some exponential formulas to simplify the value. Now on the RHS part we will use the logarithmic formula $\log m+\log n=\log mn$ and simplify the RHS part. Now we will equate the simplified LHS part with the given LHS part to get the required result.

Complete step by step answer:
Given equation is ${{\log }_{a}}x=\dfrac{3}{2}{{\log }_{a}}9+{{\log }_{a}}2$.
Considering the RHS part in the above equation, then we will have
$R.H.S=\dfrac{3}{2}{{\log }_{a}}9+{{\log }_{a}}2$
In the above equation for the value $\dfrac{3}{2}{{\log }_{a}}9$ we are going to apply the logarithmic formula $m\log n=\log {{n}^{m}}$, then the above equation is modified as
$\Rightarrow R.H.S={{\log }_{a}}{{9}^{\dfrac{3}{2}}}+{{\log }_{a}}2$
Substituting the known value $9={{3}^{2}}$ in the above equation and using the exponential rule ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$, then we will get
$\begin{align}
  & \Rightarrow R.H.S={{\log }_{a}}{{\left( {{3}^{2}} \right)}^{\dfrac{3}{2}}}+{{\log }_{a}}2 \\
 & \Rightarrow R.H.S={{\log }_{a}}{{3}^{2\times \dfrac{3}{2}}}+{{\log }_{a}}2 \\
 & \Rightarrow R.H.S={{\log }_{a}}{{3}^{3}}+{{\log }_{a}}2 \\
\end{align}$
We know that the value of ${{3}^{3}}$ is $27$. Substituting this value in the above equation, then we will have
$\Rightarrow R.H.S={{\log }_{a}}27+{{\log }_{a}}2$
Applying the logarithmic formula $\log m+\log n=\log mn$ in the above equation, then we will have
$\begin{align}
  & \Rightarrow R.H.S={{\log }_{a}}\left( 27\times 2 \right) \\
 & \Rightarrow R.H.S={{\log }_{a}}54 \\
\end{align}$
Equating the above calculated RHS with the LHS of the given equation, then we will get
$\Rightarrow {{\log }_{a}}x={{\log }_{a}}54$
In the above we can observe that the bases of the logarithm functions on both sides are the same. So, we can equate the value, hence we have
$\therefore x=54$

Note: In this problem we have the bases of all the logarithmic functions involved in the problem as the same. So, we have solved the problem very easily. So, times the bases of the logarithmic functions may not be the same, then we need to calculate the values of those functions using the logarithmic table and solve the problem.