Question

How to find the exact value of ${{\log }_{2}}\sqrt{2}$ ? \[\]

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 ${{\log }_{2}}\sqrt{2}$ argument is $\sqrt{2}$ and base is 2 . We assume ${{\log }_{2}}\sqrt{2}=y$ and use the definition of logarithm to solve for $y$. We alternatively use the logarithmic identities $m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}$,${{\log }_{b}}b=1$. 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${{\log }_{2}}\sqrt{2}$. Let us assume
\[{{\log }_{2}}\sqrt{2}=y\]
We see here that the base is $b=4$ and the argument $x=\dfrac{1}{4}$. So by definition of logarithm we have
\[\Rightarrow {{2}^{^{y}}}=\sqrt{2}\]
We know that we can write the square root as base raised to the exponent$\dfrac{1}{2}$ . So we have
\[\Rightarrow {{2}^{^{y}}}={{2}^{\dfrac{1}{2}}}\]
We equate the exponents of 2 both sides since the base at both sides is equal to have
\[\begin{align}
  & \Rightarrow y=\dfrac{1}{2} \\
 & \Rightarrow {{\log }_{2}}\sqrt{2}=-1\left( \because y={{\log }_{2}}\sqrt{2} \right) \\
\end{align}\]

Note:
We know the logarithmic identity involving power as $m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}$ . Let u s consider ${{\log }_{2}}\sqrt{2}={{\log }_{2}}{{2}^{\dfrac{1}{2}}}$. We use the identity of power for $m=\dfrac{1}{2},x=b=2$ to have;
\[{{\log }_{2}}\sqrt{2}={{\log }_{2}}{{2}^{\dfrac{1}{2}}}=\dfrac{1}{2}{{\log }_{2}}2\]
We know that when base and argument are equal we have ${{\log }_{b}}b=1$. We use this property for $b=2$ in the above step to have
\[{{\log }_{2}}\sqrt{2}={{\log }_{2}}{{2}^{\dfrac{1}{2}}}=\dfrac{1}{2}{{\log }_{2}}2=\dfrac{1}{2}\times 1=\dfrac{1}{2}\]
We note that the logarithms with base 2 are called binary logarithms.