Question

How do you evaluate $ {\log _{16}}\,\left( {\dfrac{1}{2}} \right)\, $ ?

Answer 1

Hint: To solve this equation, we have to generally use the theorem of Logarithm which is given by
When Positive real numbers and b does not equal to 1, then
 $ {\log _b}\,\left( x \right)\, = \,y $ is equal to $ x = {b^y} $

Complete step by step solution:
We know that this question can be easily solved by using property of logarithm for the given question, we have to evaluate $ {\log _{16}}\,\left( {\dfrac{1}{2}} \right)\, $
Let us assume $ {\log _{16}}\,\left( {\dfrac{1}{2}} \right)\, = \,x $
By applying the theorem which we know that
 $ {\log _b}\,\left( x \right)\, = \,y $ is equal to $ x = {b^y} $
On comparing and putting the respective value we get
 $ \dfrac{1}{2} = {\left( {16} \right)^x} $
Since, we know that we can write
 $ \dfrac{1}{2} = \,{\left( 2 \right)^{ - 1}} $
Also we know that
 $ 16 = {2^4} $
Thus on simplifying we get
 $ {2^{ - 1}} = {\left( {{2^4}} \right)^x}\, $
 $ \Rightarrow \,\,{2^{ - 1}} = {2^{4x}} $
After, comparing from both side, we get
 $ - 1 = 4x $
 $ \Rightarrow x = - \dfrac{1}{4} $
So, the value of $ {\log _{16}}\,\left( {\dfrac{1}{2}} \right)\, $ is $ \dfrac{1}{4} $
So, the correct answer is “ $ \dfrac{1}{4} $ ”.

Note: To solve the above equation we must have the basic knowledge of the basic theorem of logarithm and some basic mathematics formulae and understanding. These properties of logarithm are very useful in solving very complex equations in mathematics.