Question

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

Answer 1

Hint: We recall the definition of logarithm with base $b$ and argument $x$ as ${{b}^{y}}=x\Leftrightarrow {{\log }_{b}}x=y$. Here we are given ${{\log }_{4}}\dfrac{1}{4}$ argument is $\dfrac{1}{4}$and base is 4 . We assume ${{\log }_{4}}\dfrac{1}{4}=y$ and use the definition of logarithm to solve for $y$. We alternatively use the logarithmic identities ${{\log }_{b}}m-{{\log }_{b}}n={{\log }_{b}}\left( \dfrac{m}{n} \right)$ and ${{\log }_{b}}b=1$ to evaluate.

Complete step by step answer:
We know that the logarithm is the inverse operation 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$x$, which means if ${{b}^{y}}=x$ then the logarithm denoted as log and calculated as
\[{{\log }_{b}}x=y\]
Here $x$ is called the argument of the logarithm. The argument of the logarithm of the logarithm is always positive $\left( x>0 \right)$ and the base is also positive and never equal to 1 $\left( b>0,b\ne 1 \right)$. We are asked in the question to evaluate the value of the logarithm${{\log }_{4}}\left( \dfrac{1}{4} \right)$. Let us assume
\[{{\log }_{4}}\dfrac{1}{4}=y\]
We see here that the base is $b=4$ and the argument$x=\dfrac{1}{4}$. So by definition of logarithm we have
\[\Rightarrow {{4}^{^{y}}}=\dfrac{1}{4}\]
We know that we can write the reciprocal of any non-zero integer $a$ as ${{a}^{-1}}=\dfrac{1}{a}$. So we write for $a=4$ the reciprocal will be ${{4}^{-1}}=\dfrac{1}{4}$. We use this in the above step to have;
\[\Rightarrow {{4}^{^{y}}}={{4}^{-1}}\]
We equate the exponents of 4 both sides since the base at both sides is equal to have
\[\begin{align}
  & \Rightarrow y=-1 \\
 & \Rightarrow {{\log }_{4}}\left( \dfrac{1}{4} \right)=-1\left( \because y={{\log }_{4}}\left( \dfrac{1}{4} \right) \right) \\
\end{align}\]

Note:
We know the logarithmic identity involving quotient as ${{\log }_{b}}m-{{\log }_{b}}n={{\log }_{b}}\left( \dfrac{m}{n} \right)$ . We use the logarithmic identity of quotient to have;
\[{{\log }_{4}}\left( \dfrac{1}{4} \right)={{\log }_{4}}1-{{\log }_{4}}4\]
We know that when base and argument are equal we have ${{\log }_{b}}b=1$. We also know that the logarithm of 1 for any valid base is 0 that is ${{\log }_{b}}1=0$. We use these properties for $b=4$ in the above step to have
\[{{\log }_{4}}\left( \dfrac{1}{4} \right)={{\log }_{4}}1-{{\log }_{4}}4=0-1=-1\]