# Find the logarithms of 125 to base$5\sqrt 5 $, and 0.25 to base 4.

Answer

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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.

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