
How do you evaluate $ {{\log }_{81}}\dfrac{1}{3} $ ?
Answer
563.4k+ views
Hint: To evaluate $ {{\log }_{81}}\dfrac{1}{3} $ , we first of all convert $ \dfrac{1}{3} $ into $ {{3}^{-1}} $ and also will write 81 in terms of 3 raise to power 4. After that, we are going to use the following properties of logarithm to simplify the given expression which are shown as: $ {{\log }_{b}}a=\dfrac{{{\log }_{10}}a}{{{\log }_{10}}b} $ and $ {{\log }_{b}}{{a}^{m}}=m{{\log }_{b}}a $ .
Complete step by step answer:
In the above problem, we are asked to evaluate the following logarithm:
$ {{\log }_{81}}\dfrac{1}{3} $
In the above logarithm, we can write $ \dfrac{1}{3} $ as $ {{3}^{-1}} $ then the above expression will look as:
$ {{\log }_{81}}{{\left( 3 \right)}^{-1}} $
Now, as you can see in the above logarithm, that base of the logarithm is some power of 3 so we can write 81 as $ {{3}^{4}} $ . Substituting $ {{3}^{4}} $ in place of 81 we get,
$ {{\log }_{{{3}^{4}}}}{{3}^{-1}} $
We know the property of logarithm that:
$ {{\log }_{b}}a=\dfrac{{{\log }_{10}}a}{{{\log }_{10}}b} $
Using the above property of logarithm in $ {{\log }_{{{3}^{4}}}}{{3}^{-1}} $ we get,
$ \dfrac{{{\log }_{10}}{{3}^{-1}}}{{{\log }_{10}}{{3}^{4}}} $
We also know the exponent property of the logarithm which is equal to:
$ {{\log }_{b}}{{a}^{m}}=m{{\log }_{b}}a $
Using the above logarithm property in simplifying $ \dfrac{{{\log }_{10}}{{3}^{-1}}}{{{\log }_{10}}{{3}^{4}}} $ we get,
$ \dfrac{\left( -1 \right){{\log }_{10}}3}{\left( 4 \right){{\log }_{10}}3} $
In the above expression $ {{\log }_{10}}3 $ will be cancelled out from the numerator and the denominator and we are left with:
$ -\dfrac{1}{4} $
Hence, we have evaluated the given logarithmic expression to $ -\dfrac{1}{4} $ .
Note:
To solve the above problem, you should know the value of 3 raised to the power of 4 so remembering the power of 2 and 3 till 10 will help you in solving logarithm problems. Also, you should have a clear understanding of the logarithm properties like:
$ {{\log }_{b}}a=\dfrac{{{\log }_{10}}a}{{{\log }_{10}}b} $ and $ {{\log }_{b}}{{a}^{m}}=m{{\log }_{b}}a $
Failing the knowledge of the above concepts will stop you from solving the above problem so make sure you have a sound grasp on the logarithm properties and power of 2 and 3.
There is a tip in solving the logarithm problems. It will be better if you can remember the power of 4 and 5 also. Remembering these powers will save you time in the examination.
Complete step by step answer:
In the above problem, we are asked to evaluate the following logarithm:
$ {{\log }_{81}}\dfrac{1}{3} $
In the above logarithm, we can write $ \dfrac{1}{3} $ as $ {{3}^{-1}} $ then the above expression will look as:
$ {{\log }_{81}}{{\left( 3 \right)}^{-1}} $
Now, as you can see in the above logarithm, that base of the logarithm is some power of 3 so we can write 81 as $ {{3}^{4}} $ . Substituting $ {{3}^{4}} $ in place of 81 we get,
$ {{\log }_{{{3}^{4}}}}{{3}^{-1}} $
We know the property of logarithm that:
$ {{\log }_{b}}a=\dfrac{{{\log }_{10}}a}{{{\log }_{10}}b} $
Using the above property of logarithm in $ {{\log }_{{{3}^{4}}}}{{3}^{-1}} $ we get,
$ \dfrac{{{\log }_{10}}{{3}^{-1}}}{{{\log }_{10}}{{3}^{4}}} $
We also know the exponent property of the logarithm which is equal to:
$ {{\log }_{b}}{{a}^{m}}=m{{\log }_{b}}a $
Using the above logarithm property in simplifying $ \dfrac{{{\log }_{10}}{{3}^{-1}}}{{{\log }_{10}}{{3}^{4}}} $ we get,
$ \dfrac{\left( -1 \right){{\log }_{10}}3}{\left( 4 \right){{\log }_{10}}3} $
In the above expression $ {{\log }_{10}}3 $ will be cancelled out from the numerator and the denominator and we are left with:
$ -\dfrac{1}{4} $
Hence, we have evaluated the given logarithmic expression to $ -\dfrac{1}{4} $ .
Note:
To solve the above problem, you should know the value of 3 raised to the power of 4 so remembering the power of 2 and 3 till 10 will help you in solving logarithm problems. Also, you should have a clear understanding of the logarithm properties like:
$ {{\log }_{b}}a=\dfrac{{{\log }_{10}}a}{{{\log }_{10}}b} $ and $ {{\log }_{b}}{{a}^{m}}=m{{\log }_{b}}a $
Failing the knowledge of the above concepts will stop you from solving the above problem so make sure you have a sound grasp on the logarithm properties and power of 2 and 3.
There is a tip in solving the logarithm problems. It will be better if you can remember the power of 4 and 5 also. Remembering these powers will save you time in the examination.
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