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

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.