Question

Given: $\log 2 = 0.3010$ and $\log 3 = 0.4771$, then find the value of $\log 25$.

Answer 1

Here we will logarithmic properties to find the value of $\log 25$. First we will write the 25 which is in the log function as a fraction. Then we will simplify it using the logarithmic properties. We will simplify it further and substitute the given logarithmic values. Then we solve the equation to get the required value.

Complete step-by-step answer:
First, we will write the $\log 25$ in the modified form by writing the number 25 in a different form. Therefore, we get
$\log 25 = \log \dfrac{{100}}{4}$
Now we will use the property of the logarithmic function i.e. $\log a - \log b = \log \dfrac{a}{b}$.
Therefore, by using this property, we get
$\Rightarrow \log 25 = \log 100 - \log 4$
We know that number 100 is the square of number 10 and number 4 is the square of number 2. Now we will write this in the above equation, we get
$\Rightarrow \log 25 = \log {10^2} - \log {2^2}$
Now by using the property $\log {a^b} = b\log a$, we get
$\Rightarrow \log 25 = 2\log 10 - 2\log 2$
We know that the value of $\log 2$ is given in the question and we know that $\log 10 = 1$. Therefore, we get
$\Rightarrow \log 25 = 2\left( 1 \right) - 2\left( {0.3010} \right)$
Now we will solve the above equation to get the value of $\log 25$. Therefore, we get
$\Rightarrow \log 25 = 2 - 0.6020$
$\Rightarrow \log 25 = 1.3980$
Hence, the value of $\log 25$ is equal to $1.398$.

Note: Here in this question, we have to modify the number in the main equation according to the given values of log in the question. We should know that the value inside the log function should never be zero or negative it should always be greater than zero. Always remember that the value of the $\log 10$ is equal to 1. We should simplify the equation carefully and apply the properties of the log function accurately.
Some of the basic properties of the log functions are listed below.
1.$\log a + \log b = \log ab$
2.$\log {a^b} = b\log a$
3.$\log a - \log b = \log \dfrac{a}{b}$
4.${\log _a}b = \dfrac{{\log b}}{{\log a}}$