Question

How do you write ${{a}^{2}}=36$ in log form?

Answer 1

Hint: For writing the given equation in the log form, we have to take the logarithm on both the sides of the equation. On taking the logarithm, we will get ${{\log }_{a}}\left( {{a}^{2}} \right)={{\log }_{a}}\left( 36 \right)$. By using the logarithm property $\log \left( {{x}^{m}} \right)=m\log x$, we will be able to simplify the obtained equation. Then on further using the other properties of the logarithm function, we will be able to solve the equation.

Complete step by step answer:
The equation given to us in the above question is written as
$\Rightarrow {{a}^{2}}=36$
Taking the logarithm with the base equal to a on both the sides of the above equation, we get
$\Rightarrow {{\log }_{a}}\left( {{a}^{2}} \right)={{\log }_{a}}\left( 36 \right)$
By the properties of the logarithm function, we know that $\log \left( {{x}^{m}} \right)=m\log x$. Therefore, we can simplify the LHS of the above equation as $2{{\log }_{a}}\left( a \right)$ so that we can write the above equation as
$\Rightarrow 2{{\log }_{a}}\left( a \right)={{\log }_{a}}\left( 36 \right)$
On dividing both the sides of the above equation by $2$, we get
$\Rightarrow {{\log }_{a}}\left( a \right)=\dfrac{1}{2}{{\log }_{a}}\left( 36 \right)$
Now, using the logarithm property given by $m\log x=\log {{x}^{m}}$, we can simplify the RHS of the above equation as ${{\log }_{a}}\left( {{36}^{\dfrac{1}{2}}} \right)$ so that we can write the above equation as
$\begin{align}
  & \Rightarrow {{\log }_{a}}\left( a \right)={{\log }_{a}}\left( {{36}^{\dfrac{1}{2}}} \right) \\
 & \Rightarrow {{\log }_{a}}\left( a \right)={{\log }_{a}}\left( \sqrt{36} \right) \\
\end{align}$
We know that $\sqrt{36}=6$. On putting this in the above equation, we get
\[\Rightarrow {{\log }_{a}}\left( a \right)={{\log }_{a}}\left( 6 \right)\]
On subtracting ${{\log }_{a}}\left( 6 \right)$ from both the sides, we get
\[\begin{align}
  & \Rightarrow {{\log }_{a}}\left( a \right)-{{\log }_{a}}\left( 6 \right)={{\log }_{a}}\left( 6 \right)-{{\log }_{a}}\left( 6 \right) \\
 & \Rightarrow {{\log }_{a}}\left( a \right)-{{\log }_{a}}\left( 6 \right)=0 \\
\end{align}\]
From the logarithm property $\log a-\log b=\log \left( \dfrac{a}{b} \right)$, we can write the above equation as
$\Rightarrow {{\log }_{a}}\left( \dfrac{a}{6} \right)=0$
Now, we know that the logarithm function gives zero for the argument equal to one. Therefore, we can equate the argument of the logarithm function in the above equation to one so that we get
$\begin{align}
  & \Rightarrow \dfrac{a}{6}=1 \\
 & \Rightarrow a=6 \\
\end{align}$

Hence, we have written the given equation in the log form as ${{\log }_{a}}\left( \dfrac{a}{6} \right)=0$ and obtained the solution as $a=6$.

Note: On solving the given equation ${{a}^{2}}=36$, the values for a which we will get are \[a=-6\] and $a=6$. But in the above solution we obtained only $a=6$. This is due to the fact that the logarithm function can take only positive numbers as its argument.