Question

How do you evaluate ${\log _{64}}\left( {\dfrac{1}{8}} \right)$ ?

Answer 1

Hint : In this question we need to evaluate ${\log _{64}}\left( {\dfrac{1}{8}} \right)$ . Here, we will consider the required value as $x$ . We know that the log form and exponential form are interchangeable. Hence, we will rewrite the term from ${\log _a}b = c$ to ${a^c} = b$ . Then rewrite the bases and evaluate using the exponential formulas and we will determine the value of $x$ which is the required solution.

Complete step-by-step answer :
We need to evaluate ${\log _{64}}\left( {\dfrac{1}{8}} \right)$ .
As we know the log form and exponential form are interchangeable, we have,
${\log _a}b = c$
This can be written as,
${a^c} = b$
Thus, we can write ${\log _{64}}\left( {\dfrac{1}{8}} \right) = x$ as ${64^x} = \left( {\dfrac{1}{8}} \right)$ .
${64^x} = \left( {\dfrac{1}{8}} \right)$ $\to \left( 1 \right)$
Now, let us write this in exponential form,
${\left( {{8^2}} \right)^x} = {8^{ - 1}}$
We know that ${\left( {{a^m}} \right)^n} = {a^{mn}}$ , thus we have,
${8^{2x}} = {8^{ - 1}}$
When the bases are the same, then we can equate the powers.
$2x = - 1$
$x = - \dfrac{1}{2}$
Therefore, substituting the value of $x$ in equation $\left( 1 \right)$ ,
${64^{ - \dfrac{1}{2}}} = \left( {\dfrac{1}{8}} \right)$
Hence, ${\log _{64}}\left( {\dfrac{1}{8}} \right) = - \dfrac{1}{2}$ .
So, the correct answer is “ $- \dfrac{1}{2}$ ”.

Note : In this question it is important to note here that if the base is not given then, considering the base of the log as $10$ is the most common method used for solving these types of questions. We will also consider $e$ as the base because exponential form is the inverse of logarithm. Logarithms are the opposite of exponentials, just as subtraction is the opposite of addition and multiplication i.e., a logarithm says how many of one number to multiply to get another number and the exponent of a number says how many times to use the number in a multiplication. And, from the definition of logarithm, if $a$ and $b$ are positive real numbers and $a \ne 1$ , then ${\log _e}x = 4$ is equivalent to ${a^c} = b$ . If we can remember this relation, then we will not have too much trouble with logarithms.