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.