
How do you evaluate the expression \[{{\log }_{2}}\left( \dfrac{1}{32} \right)\]?
Answer
492k+ views
Hint: This type of problem is based on the concept of properties of logarithm. First, we have to consider the given expression. Simplify the \[\dfrac{1}{32}\] to \[\dfrac{1}{{{2}^{5}}}\]. Then, substitute \[\dfrac{1}{{{2}^{5}}}\] in the given function. And convert \[\dfrac{1}{{{2}^{5}}}\] to \[{{2}^{-5}}\] using the property \[\dfrac{1}{a}={{a}^{-1}}\]. Now, using the rule of logarithm, that is \[\log {{a}^{n}}=n\log a\], simplify the expression further. Using the property of logarithm, that is \[{{\log }_{a}}\left( a \right)=1\], 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 \[{{\log }_{2}}\left( \dfrac{1}{32} \right)\].
We have been given the expression \[{{\log }_{2}}\left( \dfrac{1}{32} \right)\]. -----------(1)
Let us first consider \[\dfrac{1}{32}\].
We know that \[32={{2}^{5}}\]. On substituting in this in \[\dfrac{1}{32}\], we get
\[\dfrac{1}{32}=\dfrac{1}{{{2}^{5}}}\].
Using the property of division, that is \[\dfrac{1}{a}={{a}^{-1}}\], we get
\[\Rightarrow \dfrac{1}{32}={{2}^{-5}}\]
On substituting in expression (1), we get
\[{{\log }_{2}}\left( \dfrac{1}{32} \right)={{\log }_{2}}\left( {{2}^{-5}} \right)\] -------------(2)
Let us now use the rule of logarithm, that is \[\log {{a}^{n}}=n\log a\].
Here, a=2 and n=-5.
Therefore, we get
\[{{\log }_{2}}\left( \dfrac{1}{32} \right)=-5{{\log }_{2}}\left( 2 \right)\]
Here, the base of logarithm is 2 and the function on which the log is defined is also 2.
We know that \[{{\log }_{a}}\left( a \right)=1\] where ‘a’ is a constant.
Using this property of algorithm, we get
\[{{\log }_{2}}\left( \dfrac{1}{32} \right)=-5\times 1\]
On further simplifications, we get
\[{{\log }_{2}}\left( \dfrac{1}{32} \right)=-5\]
Hence the value of the expression \[{{\log }_{2}}\left( \dfrac{1}{32} \right)\] 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 \[{{\log }_{2}}\left( 2 \right)\] to be 0 which will lead to an incorrect answer.
Complete step-by-step solution:
According to the question, we are asked to evaluate the expression \[{{\log }_{2}}\left( \dfrac{1}{32} \right)\].
We have been given the expression \[{{\log }_{2}}\left( \dfrac{1}{32} \right)\]. -----------(1)
Let us first consider \[\dfrac{1}{32}\].
We know that \[32={{2}^{5}}\]. On substituting in this in \[\dfrac{1}{32}\], we get
\[\dfrac{1}{32}=\dfrac{1}{{{2}^{5}}}\].
Using the property of division, that is \[\dfrac{1}{a}={{a}^{-1}}\], we get
\[\Rightarrow \dfrac{1}{32}={{2}^{-5}}\]
On substituting in expression (1), we get
\[{{\log }_{2}}\left( \dfrac{1}{32} \right)={{\log }_{2}}\left( {{2}^{-5}} \right)\] -------------(2)
Let us now use the rule of logarithm, that is \[\log {{a}^{n}}=n\log a\].
Here, a=2 and n=-5.
Therefore, we get
\[{{\log }_{2}}\left( \dfrac{1}{32} \right)=-5{{\log }_{2}}\left( 2 \right)\]
Here, the base of logarithm is 2 and the function on which the log is defined is also 2.
We know that \[{{\log }_{a}}\left( a \right)=1\] where ‘a’ is a constant.
Using this property of algorithm, we get
\[{{\log }_{2}}\left( \dfrac{1}{32} \right)=-5\times 1\]
On further simplifications, we get
\[{{\log }_{2}}\left( \dfrac{1}{32} \right)=-5\]
Hence the value of the expression \[{{\log }_{2}}\left( \dfrac{1}{32} \right)\] 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 \[{{\log }_{2}}\left( 2 \right)\] to be 0 which will lead to an incorrect answer.
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