Question

What is the logarithm of $100$ divided by $25$ ?

Answer 1

Hint: Here we are asked to find the logarithm of hundred divided by twenty-five. This problem can be done in two ways, we will be doing it both ways. The first way is simplifying the hundred divided by 25 by taking the logarithm. Another way is taking the logarithm for a hundred then dividing it by twenty-five.

Complete step-by-step answer:
We aim to find the logarithm of hundred divided by twenty-five.
This problem can be done in two ways. It is given that what is the logarithm of hundred divided by twenty-five. Firstly, we take it as a logarithm of \[\dfrac{{100}}{{25}}\] .
Consider the given problem as $\log \left( {\dfrac{{100}}{{25}}} \right)$
Let us simplify the inner fraction.
\[ = \log \left( 4 \right)\]
Now to make it simpler let us break four into an exponent term.
$ = \log {\left( 2 \right)^2}$
We know that \[\log {\left( a \right)^b} = b\log \left( a \right)\] here $a = 2$ and $b = 2$ (power) so the above can be written as
$ = 2\log \left( 2 \right)$
From the logarithm table, we get \[\log \left( 2 \right) \approx 0.30103\] substituting this in the above we get
\[ = 2 \times 0.30103\]
On simplifying this we get
\[ = \log \left( {\dfrac{{100}}{{25}}} \right) \approx 0.60206\]
On the other hand, let us consider the given problem as \[\dfrac{{\log \left( {100} \right)}}{{25}}\] .
Let us first break the hundred into an exponent term for our convenience.
$ = \dfrac{{\log {{\left( {10} \right)}^2}}}{{25}}$
Again, by using the property that \[\log {\left( a \right)^b} = b\log \left( a \right)\] we get
\[ = \dfrac{{2\log \left( {10} \right)}}{{25}}\]
From the logarithm table, we get $\log \left( {10} \right) = 1$ substituting it in the above we get
$ = \dfrac{{2 \times 1}}{{25}}$
On simplifying the above, we get
\[ = \dfrac{{\log \left( {100} \right)}}{{25}} = 0.08\]
Thus, we have found the answer to this problem in two ways.

Note: The logarithm is nothing but the inverse function of the exponent. The formula of a logarithm is given by ${\log _b}\left( {{b^x}} \right) = x$ here $b$ is the logarithm base. This problem can be taken in two ways but the first way of calculation can be preferred as it has a good approximation of \[\log \left( 2 \right)\] which is more precise than the second calculation.