Question

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

Answer 1

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 $5\sqrt{5}\text{ = 5}\times {\text{5}}^{{\displaystyle \frac{1}{2}}}=5^{1+{\displaystyle \frac{1}{2}}}=5^{{\displaystyle \frac{3}{2}}}$
Now using the property of logarithm of ${\log }_{b^p}a={\displaystyle \frac{1}{p}}{\log }_{b}a$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 ${\displaystyle \frac{1}{{\displaystyle \frac{1}{3}}}}\times 3{\log }_{5}5$
$\Rightarrow 9\times {\log }_{5}5$
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 ${\log }_{2^2}{\left(0.5\right)}^2$
$\Rightarrow {\log }_{2^2}{\left({\displaystyle \frac{1}{2}}\right)}^2$
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
$\Rightarrow {\displaystyle \frac{1}{2}}\times -2{\log }_{2}2$
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.