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Question

Question: How do you evaluate the expression \[{{\log }_{2}}\left( \dfrac{1}{32} \right)\]?...

How do you evaluate the expression log2(132){{\log }_{2}}\left( \dfrac{1}{32} \right)?

Explanation

Solution

This type of problem is based on the concept of properties of logarithm. First, we have to consider the given expression. Simplify the 132\dfrac{1}{32} to 125\dfrac{1}{{{2}^{5}}}. Then, substitute 125\dfrac{1}{{{2}^{5}}} in the given function. And convert 125\dfrac{1}{{{2}^{5}}} to 25{{2}^{-5}} using the property 1a=a1\dfrac{1}{a}={{a}^{-1}}. Now, using the rule of logarithm, that is logan=nloga\log {{a}^{n}}=n\log a, simplify the expression further. Using the property of logarithm, that is loga(a)=1{{\log }_{a}}\left( a \right)=1, we get the value of the expression equal to -5 which is the required answer.

Complete step-by-step solution:
According to the question, we are asked to evaluate the expression log2(132){{\log }_{2}}\left( \dfrac{1}{32} \right).
We have been given the expression log2(132){{\log }_{2}}\left( \dfrac{1}{32} \right). -----------(1)
Let us first consider 132\dfrac{1}{32}.
We know that 32=2532={{2}^{5}}. On substituting in this in 132\dfrac{1}{32}, we get
132=125\dfrac{1}{32}=\dfrac{1}{{{2}^{5}}}.
Using the property of division, that is 1a=a1\dfrac{1}{a}={{a}^{-1}}, we get
132=25\Rightarrow \dfrac{1}{32}={{2}^{-5}}
On substituting in expression (1), we get
log2(132)=log2(25){{\log }_{2}}\left( \dfrac{1}{32} \right)={{\log }_{2}}\left( {{2}^{-5}} \right) -------------(2)
Let us now use the rule of logarithm, that is logan=nloga\log {{a}^{n}}=n\log a.
Here, a=2 and n=-5.
Therefore, we get
log2(132)=5log2(2){{\log }_{2}}\left( \dfrac{1}{32} \right)=-5{{\log }_{2}}\left( 2 \right)
Here, the base of logarithm is 2 and the function on which the log is defined is also 2.
We know that loga(a)=1{{\log }_{a}}\left( a \right)=1 where ‘a’ is a constant.
Using this property of algorithm, we get
log2(132)=5×1{{\log }_{2}}\left( \dfrac{1}{32} \right)=-5\times 1
On further simplifications, we get
log2(132)=5{{\log }_{2}}\left( \dfrac{1}{32} \right)=-5
Hence the value of the expression log2(132){{\log }_{2}}\left( \dfrac{1}{32} \right) is -5.

Note: Whenever you get this type of problems, we should simplify the given expression. We should be thorough with the logarithmic properties to solve this question. We should avoid calculation mistakes based on sign conventions. Also do not consider the value of log2(2){{\log }_{2}}\left( 2 \right) to be 0 which will lead to an incorrect answer.