Question

How do you evaluate ${\log _{32}}\left( {\dfrac{1}{2}} \right)$?

Answer 1

Hint: In this question, we are given with an expression in logarithm and asked to solve it. Using the property of the logarithm mentioned below, first we will eliminate the log in the expression. Then we will solve for $y$ using the properties of exponential to get the required answer.

Formulas used:
If${a^x} = b$ then, $x = {\log _a}b$
${x^{ - y}} = \dfrac{1}{{{x^y}}}$
$\log {x^a} = a\log x$

Complete Step by Step Solution:
In this question, we are asked to solve an expression given in logarithm ${\log _{32}}\left( {\dfrac{1}{2}} \right)$.
We will be solving this by using the properties of logarithm.
First let ${\log _{32}}\left( {\dfrac{1}{2}} \right)$ be $y$.
Using the basic property or formation of logarithm that is,
If${a^x} = b$ then, $x = {\log _a}b$
Using this property of logarithm, we will eliminate log
$ \Rightarrow {\log _{32}}\left( {\dfrac{1}{2}} \right) = y$
$ \Rightarrow {32^y} = \left( {\dfrac{1}{2}} \right)$
Keeping this aside,
Now, we can also write $32$ as ${2^5}$ i.e.,$32 = {2^5}$
Replacing $32$ by ${2^5}$ we get,
$ \Rightarrow {\left( {{2^5}} \right)^y} = \left( {\dfrac{1}{2}} \right)$
Multiplying the powers inside the bracket (the property of power and exponentials)
$ \Rightarrow {2^{5y}} = \left( {\dfrac{1}{2}} \right)$
Now, since there is $2$ on both sides, we should try to rearrange it in such a way that it can get cancelled or eliminated without affecting the value of the term.
Clearly, we can change the LHS, by inverting it. In order to retain the real value of the term, we will change the sign of the power, as we know that ${x^{ - y}} = \dfrac{1}{{{x^y}}}$
\[ \Rightarrow \dfrac{1}{{{2^{ - 5y}}}} = \dfrac{1}{{{2^1}}}\]
Now both the bases are the same and LHS is equal to RHS, it means that the powers are also equal. Therefore,
$ \Rightarrow - 5y = 1$
$ \Rightarrow y = - \dfrac{1}{5}$
We know that ${\log _{32}}\left( {\dfrac{1}{2}} \right) = y$

Therefore the value of ${\log _{32}}\left( {\dfrac{1}{2}} \right)$ is $ - \dfrac{1}{5}$

Note: Alternative method:
The given question is ${\log _{32}}\left( {\dfrac{1}{2}} \right)$.
$ \Rightarrow {\log _{32}}\left( {\dfrac{1}{2}} \right)$
This can be written as,
$ \Rightarrow \dfrac{{{{\log }_x}\left( {\dfrac{1}{2}} \right)}}{{{{\log }_x}32}}$
Now we should try to rearrange it.
As we know that ${x^{ - y}} = \dfrac{1}{{{x^y}}}$
We can write,
$ \Rightarrow \dfrac{{{{\log }_x}{{\left( 2 \right)}^{ - 1}}}}{{{{\log }_x}32}}$
We can also write $32$ as ${2^5}$i.e.,$32 = {2^5}$
Replacing $32$ by ${2^5}$ we get,
$ \Rightarrow \dfrac{{{{\log }_x}{{\left( 2 \right)}^{ - 1}}}}{{{{\log }_x}{{\left( 2 \right)}^5}}}$
From the property of logarithm that is $\log {x^a} = a\log x$ , we can write,
$ \Rightarrow \dfrac{{ - {{\log }_x}\left( 2 \right)}}{{5{{\log }_x}\left( 2 \right)}}$
Cancelling the $\log 2$, we get
$ \Rightarrow \dfrac{{ - 1}}{5}$
Therefore the value of ${\log _{32}}\left( {\dfrac{1}{2}} \right)$ is $ - \dfrac{1}{5}$