Question

How do you write ${{3}^{-2}}=\dfrac{1}{9}$ in log form?

Answer 1

Hint: In this question, we have to write the equation in the form of the logarithm. Therefore, to solve this question, we will use the logarithm property, to get a simplified answer in terms of the logarithm function. As we know, the logarithm function is an exponent. It means any exponential expression can be expressed in the form of the logarithm, to get a more simplified solution. We start solving this problem by putting a log on both sides of the equation and then use the logarithm property $\log ({{a}^{b}})=b(\log a)$ on the left-hand side of the equation. Then, divide both sides by (log(3)), and in last again apply the logarithm property ${{\log }_{b}}(a)=\dfrac{\log a}{\log b}$ ,to get the simplified answer in log form.

Complete answer:
According to the question, it is given that there is an equation that does not contain any variable, but constants only.
So, in this question, we will use the logarithmic properties, to get the solution in terms of the log function.
Equation: ${{3}^{-2}}=\dfrac{1}{9}$ ---------- (1)
Now, we will put the log function on both sides in equation (1), we get
$\Rightarrow \log ({{3}^{-2}})=\log \left( \dfrac{1}{9} \right)$
So, to solve log function, we will put the logarithm property $\log ({{a}^{b}})=b(\log a)$ on the left-hand side of the above equation, we get
$\Rightarrow -2.(\log 3)=\log \left( \dfrac{1}{9} \right)$
Now, we will divide both sides by log 3, to get
$\Rightarrow -2.\left( \dfrac{\log 3}{\log 3} \right)=\dfrac{\log \left( \dfrac{1}{9} \right)}{\log 3}$
On further simplification, we get
$\Rightarrow -2=\dfrac{\log \left( \dfrac{1}{9} \right)}{\log 3}$
Now, again we will use the logarithm property ${{\log }_{b}}(a)=\dfrac{\log a}{\log b}$ , therefore we get
$\Rightarrow -2={{\log }_{3}}\left( \dfrac{1}{9} \right)$
So, we see that we cannot further simplify the above equation.
Therefore, we write the equation ${{3}^{-2}}=\dfrac{1}{9}$ in terms of the logarithm as $-2={{\log }_{3}}\left( \dfrac{1}{9} \right)$ , which is our required answer.

Note: Always make calculations properly to avoid errors. Do keep in mind; in the first step, we put a log function on both sides of the equation. Always mention the properties that you are using while solving this problem. One of the alternative methods for solving the equation is we use the logarithm property at the beginning of the function, which is our required answer.
An alternative method:
Equation: ${{3}^{-2}}=\dfrac{1}{9}$ ------------ (2)
We will apply the logarithm property ${{a}^{b}}=c\Rightarrow {{\log }_{a}}c=b$ in equation (2), which is
On comparing both the equation we get, a=3, b=-2, and c=$\dfrac{1}{9}$ , therefore we get
$\Rightarrow {{3}^{-2}}=\dfrac{1}{9}$
$\Rightarrow {{\log }_{3}}\left( \dfrac{1}{9} \right)=-2$ which is our required answer.