Question

If $a=\log 12,b=\log 21,c=\log 11$ and $d=\log 22$ then $\log \left({\displaystyle \frac{1}{7}}\right)$ can be expressed in the form $P\left(a-b\right)+Q\left(c-d\right)$ where $P$ and $Q$ are integers then the value of $\left(7P-Q\right)$ equals:
A) $5$
B) $9$
C) $13$
D) $15$

Answer 1

Hint: The given values are expressed in the logarithms so we will use the properties of logarithms to simplify the given expression and then we will equal it with the given condition then we will determine the required constants. Finally, we will substitute the given values in the required equations.

Complete step-by-step answer:
The values of different variables are given in terms of different logarithms.
It is given by $a=\log 12,b=\log 21,c=\log 11$ and $d=\log 22$ .
For any two numbers $a$ and $b$ the following logarithm properties are very important in order to solve the given problem.
$\log a+\log b=\log ab$
Similarly,
$\log a-\log b=\log \left({\displaystyle \frac{a}{b}}\right)$
Also, the following property will also hold:
$m\log a=\log a^m$
We will start with the given expression.
We will substitute the given values in the expression.
$\Rightarrow$ $P\left(a-b\right)+Q\left(c-d\right)=P\left(\log 12-\log 21\right)+Q\left(\log 11-\log 22\right)$
Now we will use the rules of logarithms to simplify the brackets.
$\Rightarrow$ $P\left(a-b\right)+Q\left(c-d\right)=P\left(\log {\displaystyle \frac{4}{7}}\right)+Q\left(\log {\displaystyle \frac{1}{2}}\right)$
Now again we will expand the logs and write the given example as follows:
$\Rightarrow$ $P\left(a-b\right)+Q\left(c-d\right)=P\left(\log {\displaystyle \frac{4}{7}}\right)+Q\log 1-Q\log 2$
Using the last property of logarithms, we can write:
$\Rightarrow$ $P\left(a-b\right)+Q\left(c-d\right)=P\left(\log {\displaystyle \frac{4}{7}}\right)-Q\log 2$
We have also given that:
$\Rightarrow$ $P\left(a-b\right)+Q\left(c-d\right)=\left(\log {\displaystyle \frac{1}{7}}\right)$
From the equality we have established, we can write:
$\Rightarrow$ $$P\left(\log {\displaystyle \frac{4}{7}}\right)-Q\log 2=\left(\log {\displaystyle \frac{1}{7}}\right)$$
Now using the same properties, we can write the following:
$\Rightarrow$ $\log {\displaystyle \frac{4}{7}}=\log 4+\log {\displaystyle \frac{1}{7}}$
Therefore, the above equation becomes:
$\Rightarrow$ $$P\left(2\log 2+\log {\displaystyle \frac{1}{7}}\right)-Q\log 2=\left(\log {\displaystyle \frac{1}{7}}\right)$$
We will expand the bracket and write it as follows:
$\Rightarrow$ $$2P\log 2+P\log {\displaystyle \frac{1}{7}}-Q\log 2=\left(\log {\displaystyle \frac{1}{7}}\right)$$
We can rearrange the equation as follows:
$\Rightarrow$ $$\left(2P-Q\right)\log 2+P\log {\displaystyle \frac{1}{7}}=\left(\log {\displaystyle \frac{1}{7}}\right)$$
Now comparing the right-hand side and left-hand side we get:
$\Rightarrow$ $2P-Q=0$ and $P=1$.
Therefore, we get:
$\Rightarrow$ $2\left(1\right)-Q=0\Rightarrow Q=2$
Now we have found all the constants that are required.
So, we can find the value of the expression that is asked.
$\Rightarrow$ $\left(7P-Q\right)=7\left(1\right)-2\Rightarrow \left(7P-Q\right)=5$
Hence, the correct option is A.

Note: Here the given expressions are very crucial to solve the problem. We had to use the appropriate properties of logarithms in order to simplify the expression. After equating the simplified equation with the given value there is nothing much left to do than the substitution.