Question

How do you evaluate $ {\log _2}\left( {\dfrac{1}{4}} \right) $ ?

Answer 1

Hint: : In order to determine the value of the above question ,first split the above logarithm with base 2 using identity $ {\log _a}(b) = \dfrac{{\ln (b)}}{{\ln (a)}} $ and then rewrite the logarithm using identity $ \ln \left( {\dfrac{a}{b}} \right) = \ln (a) - \ln (b) $ to find the required result.
Formula:
 $
  n\log m = \log {m^n} \\
  {\log _a}(b) = \dfrac{{\ln (b)}}{{\ln (a)}} \\
  \ln \left( {\dfrac{a}{b}} \right) = \ln (a) - \ln (b) \\
  $

Complete step-by-step answer:
To solve the given question, we must know the properties of logarithms and with the help of them we are going to rewrite our question.
But we also need to know that the base of the natural logarithm is 10 and we have to convert this question in the form of natural logarithm as currently the base of the logarithm is 10.
First, we are going to rewrite the number with the help of the following properties of natural logarithms.
 $ {\log _a}(b) = \dfrac{{\ln (b)}}{{\ln (a)}} $
So,
 $
   = {\log _2}\left( {\dfrac{1}{4}} \right) \\
   = \dfrac{{\ln (\dfrac{1}{4})}}{{\ln (2)}} \;
  $
Using natural logarithm property $ \ln \left( {\dfrac{a}{b}} \right) = \ln (a) - \ln (b) $
 $
   = \dfrac{{\ln (1) - \ln (4)}}{{\ln (2)}} \\
   = \dfrac{{\ln (1) - \ln ({2^2})}}{{\ln (2)}} \;
  $
Value of $ \ln (1) = 0 $ and using property $ \log {m^n} = n\log m $
\[
   = \dfrac{0}{{\ln (2)}} - \dfrac{{2\ln (2)}}{{\ln (2)}} \\
   = 0 - 2(1) \\
   = - 2 \;
 \]
Therefore, value of $ {\log _2}\left( {\dfrac{1}{4}} \right) $ is equal to \[ - 2\]
So, the correct answer is “-2”.

Note: 1.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values.
 $ {\log _b}(mn) = {\log _b}(m) + {\log _b}(n) $
2. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values.
 $ {\log _b}\left( {\dfrac{m}{n}} \right) = {\log _b}(m) - {\log _b}(n) $
3. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
 $ n\log m = \log {m^n} $
4. The above guidelines work just if the bases are the equivalent. For example, the expression $ {\log _d}(m) + {\log _b}(n) $ can't be improved, on the grounds that the bases (the "d" and the "b") are not the equivalent.