Question

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

Answer 1

Hint:As we know that the logarithm is the inverse function to exponentiation. That means the logarithm of a given number $x$ is the exponent to which another fixed number, the base b, must be raised , to produce that number $x$. As per the definition of a logarithm ${\log _a}b = c$ which gives that ${a^c} = b$ . Here in the above expression the base is $16$. And we also have to assume that if no base $b$ is written then the base is always 10. This is an example of base ten logarithm because $10$ is the number that is raised to a power.

Complete step by step answer:
As per the given question we have ${\log _{16}}\left( {\dfrac{1}{4}} \right)$ . We know that if ${\log _a}b = x$, then ${a^x} = b$.Let us take ${\log _{16}}\left( {\dfrac{1}{4}} \right) = x$. BY applying the above logarithm identity we have ${16^x} = \dfrac{1}{4}$. We now have to solve for $x$by putting everything in the same base, we get: ${({2^4})^x} = \dfrac{1}{{{2^2}}}$.

There is one rule of exponent that if there is ${({a^b})^m}$then it equals ${a^{b*m}}$. Also $\dfrac{1}{a}$is the reciprocal of $a$. Therefore it can be written as ${a^{ - 1}}$. So by applying these identities we have: ${2^{4x}} = {2^{ - 2}}$ . Now we know that if the base of the powers are the same then the base gets cancelled and only the powers are solved. i.e.
$4x = - 2 \\
\Rightarrow x = \dfrac{{ - 2}}{4}$.
$\therefore x = - \dfrac{1}{2}$.

Hence the value of ${\log _{16}}\left( {\dfrac{1}{4}} \right) = - \dfrac{1}{2}$.

Note:We should always be careful while solving logarithm formulas and before solving this kind of problems we should know all the rules of logarithm and exponentiation. We have to keep in mind that when a logarithm is written without any base, like this: $\log 100$ then this usually means that the base is already there which is $10$ . It is called a common logarithm or decadic logarithm, is the logarithm to the base $10$ . One way we can approach log problems is to keep in mind that ${a^b} = c$ and ${\log _a}c = b$ .