Question

How do you evaluate ${\log _{\dfrac{1}{6}}}\left( {\dfrac{1}{{36}}} \right)$?

Answer 1

Hint: We will, first of all, use the identity that: ${\log _b}a = \dfrac{{\log a}}{{\log b}}$. Then, we will use the fact that $\log {a^b} = b\log a$ and simplify the numerator and denominator and cancel out the common terms.

Complete step by step solution:
We are given that we are required to evaluate ${\log _{\dfrac{1}{6}}}\left( {\dfrac{1}{{36}}} \right)$.
Since, we know that we have a fact given by the following expression:-
$ \Rightarrow {\log _b}a = \dfrac{{\log a}}{{\log b}}$
Replacing b by $\dfrac{1}{6}$ and a by $\dfrac{1}{{36}}$, we will then obtain the following equation with us:-
$ \Rightarrow {\log _{\dfrac{1}{6}}}\left( {\dfrac{1}{{36}}} \right) = \dfrac{{\log \left( {\dfrac{1}{{36}}} \right)}}{{\log \left( {\dfrac{1}{6}} \right)}}$
We can write this expression as following expression:-
$ \Rightarrow {\log _{\dfrac{1}{6}}}\left( {\dfrac{1}{{36}}} \right) = \dfrac{{\log {{36}^{ - 1}}}}{{\log {6^{ - 1}}}}$
We can write this expression as following expression:-
$ \Rightarrow {\log _{\dfrac{1}{6}}}\left( {\dfrac{1}{{36}}} \right) = \dfrac{{\log {{\left( {{6^2}} \right)}^{ - 1}}}}{{\log {6^{ - 1}}}}$
Now, we will use the fact that ${x^a}.{x^b} = {x^{a + b}}$. So, we will get:-
$ \Rightarrow {\log _{\dfrac{1}{6}}}\left( {\dfrac{1}{{36}}} \right) = \dfrac{{\log {6^{ - 2}}}}{{\log {6^{ - 1}}}}$
Now, we will use the fact that $\log {a^b} = b\log a$. Then, just by replacing a by 36 and 6 and b by – 1, we will then obtain the following equation with us:-
$ \Rightarrow {\log _{\dfrac{1}{6}}}\left( {\dfrac{1}{{36}}} \right) = \dfrac{{ - 2 \times \log 6}}{{ - 1 \times \log 6}}$
Crossing – off the common – log (6) from both the numerator and denominator, we will then obtain the following equation with us:-
$ \Rightarrow {\log _{\dfrac{1}{6}}}\left( {\dfrac{1}{{36}}} \right) = \dfrac{2}{1} = 2$
Thus, we have the required answer with us.

Note:
The students must commit to memory the following facts we used in the solution given above:-
${\log _b}a = \dfrac{{\log a}}{{\log b}}$
$\log {a^b} = b\log a$
${x^a}.{x^b} = {x^{a + b}}$
This is true for all positive real numbers or in some cases real numbers as well.
The students must also keep in mind that the logarithm of 0 or any negative terms are also not defined. The logarithmic is only defined for positive values. The power may be positive, negative or zero.
Therefore, in the first formula stated above in the note, a and b both can never be equal to 0 or any negative number which can be expressed as: a, b > 0.
In the second formula stated in the note above, a can never be equal to 0 or any negative number which can be expressed as a > 0 but b can take any real value.