Find the logarithms of 125 to base$5\sqrt 5 $, and 0.25 to base 4.
Answer
384.9k+ views
Hint: Use properties of logarithms.
So we have to find the ${\log _{5\sqrt 5 }}125$and ${\log _4}0.25$
Firstly let’s calculate ${\log _{5\sqrt 5 }}125$
Now $5\sqrt 5 {\text{ = 5}} \times {{\text{5}}^{\dfrac{1}{2}}} = {5^{1 + \dfrac{1}{2}}} = {5^{\dfrac{3}{2}}}$
Now using the property of logarithm of ${\log _{{b^p}}}a = \dfrac{1}{p}{\log _b}a$and ${\log _b}{a^n} = n{\log _b}a$
The above is ${\log _{{5^{\frac{1}{3}}}}}{\left( 5 \right)^3}$which can be written as $\dfrac{1}{{\dfrac{1}{3}}} \times 3{\log _5}5$
$ \Rightarrow 9 \times {\log _5}5$
Now ${\log _5}5 = 1$
Hence ${\log _{5\sqrt 5 }}125 = 9$
Now let’s calculate for ${\log _4}0.25$
This is written as ${\log _{{2^2}}}{\left( {0.5} \right)^2}$
$ \Rightarrow {\log _{{2^2}}}{\left( {\dfrac{1}{2}} \right)^2}$
This can be written as
$ \Rightarrow {\log _{{2^2}}}{\left( 2 \right)^{ - 2}}$
Now using the property of logarithm mentioned above we can write this as
$ \Rightarrow \dfrac{1}{2} \times - 2{\log _2}2$
Now ${\log _2}2 = 1$
We get ${\log _4}0.25 = - 1$
Note: Whenever we are solving such a type of problem we just need to have a grasp of the logarithm properties that were being used above, these are some of the frequently used properties of logarithm and in most of such types of questions.
So we have to find the ${\log _{5\sqrt 5 }}125$and ${\log _4}0.25$
Firstly let’s calculate ${\log _{5\sqrt 5 }}125$
Now $5\sqrt 5 {\text{ = 5}} \times {{\text{5}}^{\dfrac{1}{2}}} = {5^{1 + \dfrac{1}{2}}} = {5^{\dfrac{3}{2}}}$
Now using the property of logarithm of ${\log _{{b^p}}}a = \dfrac{1}{p}{\log _b}a$and ${\log _b}{a^n} = n{\log _b}a$
The above is ${\log _{{5^{\frac{1}{3}}}}}{\left( 5 \right)^3}$which can be written as $\dfrac{1}{{\dfrac{1}{3}}} \times 3{\log _5}5$
$ \Rightarrow 9 \times {\log _5}5$
Now ${\log _5}5 = 1$
Hence ${\log _{5\sqrt 5 }}125 = 9$
Now let’s calculate for ${\log _4}0.25$
This is written as ${\log _{{2^2}}}{\left( {0.5} \right)^2}$
$ \Rightarrow {\log _{{2^2}}}{\left( {\dfrac{1}{2}} \right)^2}$
This can be written as
$ \Rightarrow {\log _{{2^2}}}{\left( 2 \right)^{ - 2}}$
Now using the property of logarithm mentioned above we can write this as
$ \Rightarrow \dfrac{1}{2} \times - 2{\log _2}2$
Now ${\log _2}2 = 1$
We get ${\log _4}0.25 = - 1$
Note: Whenever we are solving such a type of problem we just need to have a grasp of the logarithm properties that were being used above, these are some of the frequently used properties of logarithm and in most of such types of questions.
Recently Updated Pages
Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Scroll valve is present in a Respiratory system of class 11 biology CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

is known as the Land of the Rising Sun A Japan B Norway class 8 social science CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
