Question

How do you evaluate ${{\log }_{\dfrac{1}{9}}}\left( \dfrac{1}{81} \right)$?

Answer 1

Hint: We can solve the given question by using one of the properties for log which is $ \log \left( \dfrac{x}{y} \right)=\log x-\log y$. Here, x will be 1 and y will be 81 . Then, we have to use another property which is $ \log \left( {{a}^{x}} \right)=x\log \left( a \right)$. Finally, after all this, we can get our answer.

Complete step by step solution:
According to the problem, we are asked to evaluate the equation ${{\log }_{\dfrac{1}{9}}}\left( \dfrac{1}{81} \right)$--- ( 1 )
Therefore, by using the property of log that is $ \log \left( \dfrac{x}{y} \right)=\log x-\log y$, in equation 1, we get:
Here , x = 1 and y = 81 .
Therefore, we can get:
$ \Rightarrow {{\log }_{\dfrac{1}{9}}}\left( \dfrac{1}{81} \right)={{\log }_{\dfrac{1}{9}}}\left( 1 \right)-{{\log }_{\dfrac{1}{9}}}\left( 81 \right)$ ---- (2)
We also know that log(1) = 0. ---- (3)
And for $\log \left( 81 \right)$, we can also write this as ${{\log }_{\dfrac{1}{9}}}\left( 81 \right)={{\log }_{\dfrac{1}{9}}}\left( {{9}^{2}} \right)$. As we know that ${{9}^{2}}$ = 81

$\Rightarrow {{\log }_{\dfrac{1}{9}}}\left( 81 \right)={{\log }_{\dfrac{1}{9}}}\left( {{9}^{2}} \right)$ ------ (4)

By substituting equation 3 and equation 4 in equation 2, we get
$\Rightarrow {{\log }_{\dfrac{1}{9}}}\left( \dfrac{1}{81} \right)=0-{{\log }_{\dfrac{1}{9}}}\left( {{9}^{2}} \right)$

$\Rightarrow {{\log }_{\dfrac{1}{9}}}\left( \dfrac{1}{81} \right)=-{{\log }_{\dfrac{1}{9}}}\left( {{9}^{2}} \right)$
Now, we will use the property $ \log \left( {{a}^{x}} \right)=x\log \left( a \right)$, therefore, we get

$\Rightarrow {{\log }_{\dfrac{1}{9}}}\left( \dfrac{1}{81} \right)=-2{{\log }_{\dfrac{1}{9}}}\left( 9 \right)$ ------ (5)

But, we also know that ${{\log }_{\dfrac{1}{9}}}\left( 9 \right)$= -1. If we take this as equation (6)

$\Rightarrow {{\log }_{\dfrac{1}{9}}}\left( 9 \right)=-1$ ------ (6)

Finally, by substituting equation 6 in equation 5, we get :
$\Rightarrow {{\log }_{\dfrac{1}{9}}}\left( \dfrac{1}{81} \right)=-2(-1)$

$\Rightarrow {{\log }_{\dfrac{1}{9}}}\left( \dfrac{1}{81} \right)=2$

So, we have found that after evaluating the given equation ${{\log }_{\dfrac{1}{9}}}\left( \dfrac{1}{81} \right)$ we get the answer as 2 .
Therefore, the solution of the given equation ${{\log }_{\dfrac{1}{9}}}\left( \dfrac{1}{81} \right)$ is ${{\log }_{\dfrac{1}{9}}}\left( \dfrac{1}{81} \right)=2$.

Note: The properties of log are very important for these types of problems. To solve the given equation, we could also directly use the formula ${{\log }_{a}}\left( {{a}^{x}} \right)$ = x. Here, we can take a as $\dfrac{1}{9}
$ and x = 2. $\dfrac{1}{81}$ can be written as $\dfrac{1}{{{9}^{2}}}$. If we do this, we get the answer in just two steps.