Answer
Verified
421.2k+ 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
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE