Question

How do you evaluate ${{\log }_{\dfrac{1}{3}}}\left( \dfrac{1}{9} \right)$?

Answer 1

From the question, we have been asked to evaluate ${{\log }_{\dfrac{1}{3}}}\left( \dfrac{1}{9} \right)$. Clearly, we can observe that the given expression is in logarithmic form. To solve this, we have to use some basic formulae of logarithms and exponents.

Complete step by step answer:
Now, from the question, it had been given that ${{\log }_{\dfrac{1}{3}}}\left( \dfrac{1}{9} \right)$
Now, let us equal the given logarithmic expression to $x$.
By doing this, we get the below equation ${{\log }_{\dfrac{1}{3}}}\left( \dfrac{1}{9} \right)=x$
Now, as we have already discussed above, apply one of the basic formulas of logarithms to further simplify the problem.
If ${{\log }_{\dfrac{a}{b}}}\left( \dfrac{c}{d} \right)=x$, then ${{\left( \dfrac{a}{b} \right)}^{x}}=\left( \dfrac{c}{d} \right)$
By applying the one of the basic formula of logarithms, we get the below equation ${{\log }_{\dfrac{1}{3}}}\left( \dfrac{1}{9} \right)=x$
$\Rightarrow {{\left( \dfrac{1}{3} \right)}^{x}}=\left( \dfrac{1}{9} \right)$
We know that we can write $\left( \dfrac{1}{9} \right)$ as ${{\left( \dfrac{1}{3} \right)}^{2}}$
Now, by writing $\left( \dfrac{1}{9} \right)$ as ${{\left( \dfrac{1}{3} \right)}^{2}}$, we get the below equation ${{\left( \dfrac{1}{3} \right)}^{x}}={{\left( \dfrac{1}{3} \right)}^{2}}$
We can clearly observe that the bases of the both right-hand side of the equation and the left-hand side of the equation are equal.
From one of the basic laws of exponents, we know that if the bases are equal then we can equate their powers or indices.
By equating the powers or indices of the above equation, we get $x=2$
Therefore, we got the value of $x=2$.
Hence, the given question is evaluated.

Note:
We should be well aware of the logarithms. Also, we should be well aware of the properties of logarithms and the basic formulae of logarithms. Also, we should be well aware of the basic laws of exponents that are used to solve the given question. Also, we should be very careful while doing the calculation. This can be simply answered as ${{\log }_{\dfrac{1}{3}}}\left( \dfrac{1}{9} \right)={{\log }_{\dfrac{1}{3}}}{{\left( \dfrac{1}{3} \right)}^{2}}=2$ it’s that easy to answer.