Question

$3\log 2-4\log 3$ can be written as a single logarithm with base as $10$ as:
A. $\log \dfrac{8}{81}$
B. $\log 12$
C. $\log 81$
D. $\log 6$

Answer 1

To solve the given question, we will use the some properties of logarithm. Firstly, we will use the property $\left( a\log b=\log {{b}^{a}} \right)$. Then we will expand the numerical term. Then, we will use the other property of logarithm that is $\left( \log a-\log b=\log \dfrac{a}{b} \right)$ and will simplify it into the simplest form to get the answer.

Complete step-by-step solution:
Since, we will have the given question of logarithm as:
$\Rightarrow 3\log 2-4\log 3$
Now, we will use the property of algorithm, $\left( a\log b=\log {{b}^{a}} \right)$, in the above step and we can write it as:
$\Rightarrow \log {{2}^{3}}-\log {{3}^{4}}$
Here, we will calculate the cube of $2$using the method of three times multiplication of number to itself and will get $8$ and for ${{4}^{th}}$ power of $3$ , we will do the multiplication four times of number to itself and the resultant will be $81$. So, the next step will be as:
$\Rightarrow \log 8-\log 81$
Now, we will choose another property of the algorithm to solve the above step and the property of the algorithm that will be used is $\left( \log a-\log b=\log \dfrac{a}{b} \right)$. So, we can write the above step as:
$\Rightarrow \log \dfrac{8}{81}$
Since, we are not able to do further calculation that means this is the simplest form. Hence,$3\log 2-4\log 3$ can be written as a single logarithm with base $10$ as $\log \dfrac{8}{81}$.

Note: Here are the properties of algorithm with same base as:
A. $\log \left( xy \right)=\log x+ \log y$
B. $\log \left( \dfrac{x}{y} \right)=\log x-\log y$
C. $\log \left( {{x}^{y}} \right)=y\log x$
We should have knowledge of these laws of logarithm for solving the given types of problems. We must also know the difference between $ln$ and $log$. Here ln is a natural log and is defined for the base e but the $log$ is defined for the base 10.