Question

How do you evaluate ${\log _{\dfrac{1}{4}}}(\dfrac{1}{{16}})$ ?

Answer 1

Hint: This is a $\log$ function with a base $\dfrac{1}{4}$ . At first, we will convert this $\log$ function to the $\log$ function with base $\;10$ . Then we will use the basic formulas of $\log$ function that will make the simplification easy. Then we will simplify.

Formula used: ${\log _b}c = \dfrac{{{{\log }_a}c}}{{{{\log }_a}b}}$
${\log _a}{(b)^n} = n{\log _a}b$
$\dfrac{{ab}}{{ac}} = \dfrac{b}{c}$

Complete step-by-step solution:
We have;
${\log _{\dfrac{1}{4}}}(\dfrac{1}{{16}})$
As we know that ${\log _b}c = \dfrac{{{{\log }_a}c}}{{{{\log }_a}b}}$ ;
Let us choose the basic or standard base $\;10$ as $a$ of the formula and then we will apply this to the above term.
Now applying this formula we will get;
${\log _{\dfrac{1}{4}}}(\dfrac{1}{{16}}) = \dfrac{{{{\log }_{10}}(\dfrac{1}{{16}})}}{{{{\log }_{10}}(\dfrac{1}{4})}}$
We know that $\dfrac{1}{{16}} = {\left( {\dfrac{1}{4}} \right)^2}$ .
So the previous term can be written as ${\dfrac{{{{\log }_{10}}\left( {\dfrac{1}{4}} \right)}}{{{{\log }_{10}}\left( {\dfrac{1}{4}} \right)}}^2}$ .
Now we will use the formula ${\log _a}{(b)^n} = n{\log _a}b$ in the above term.
We will get;
$= \dfrac{{2{{\log }_{10}}\left( {\dfrac{1}{4}} \right)}}{{{{\log }_{10}}\left( {\dfrac{1}{4}} \right)}}$
Now we will consider the term ${\log _{10}}\left( {\dfrac{1}{4}} \right)$ as a single term and use this formula $\dfrac{{ab}}{{ac}} = \dfrac{b}{c}$ .
We will get;
$= \dfrac{2}{1}$
$= 2$
Then we will finally get;
${\log _{\dfrac{1}{4}}}(\dfrac{1}{{16}}) = 2$ is the required answer.

Note: We have;
${\log _{\dfrac{1}{4}}}(\dfrac{1}{{16}})$
As we know that ${\log _b}c = \dfrac{{{{\log }_a}c}}{{{{\log }_a}b}}$ ;
Let us choose the basic or standard base $\;10$ as $a$ of the formula and then we will apply this to the above term.
Now applying this formula we will get;
${\log _{\dfrac{1}{4}}}(\dfrac{1}{{16}}) = \dfrac{{{{\log }_{10}}(\dfrac{1}{{16}})}}{{{{\log }_{10}}(\dfrac{1}{4})}}$
Now we will use the formula ${\log _a}\left( {\dfrac{b}{c}} \right) = {\log _a}b - {\log _a}c$ in the numerator and the denominator.
We will get;
$= \dfrac{{{{\log }_{10}}1 - {{\log }_{10}}16}}{{{{\log }_{10}}1 - {{\log }_{10}}4}}$
As we know that ${\log _{10}}1 = 0$ , we will get;
$= \dfrac{{ - {{\log }_{10}}16}}{{ - {{\log }_{10}}4}}$
$= \dfrac{{{{\log }_{10}}16}}{{{{\log }_{10}}4}}$
We also know that $16 = {4^2}$ .
Then,
$= \dfrac{{{{\log }_{10}}{{(4)}^2}}}{{{{\log }_{10}}4}}$
Now we will use the formula ${\log _a}{(b)^n} = n{\log _a}b$ in the above term.
We will get;
$= \dfrac{{2{{\log }_{10}}4}}{{{{\log }_{10}}4}}$
$= 2$