Question

If a logarithmic equation is given as $\log \left( a-b \right)=\log a-\log b$ then find the value of a in terms of b will be?

Answer 1

Hint: Compare the right hand side of the given expression with the following property of the logarithm functions:
$\operatorname{logm}-\operatorname{logn}=\log \left( \dfrac{m}{n} \right)$
And value of $\log$of any two terms will be equal if both the terms will be equal i.e. if $\operatorname{logm}=\operatorname{logn}$, then m = n

Complete step-by-step solution -
Given expression in the question is
$\log \left( a-b \right)=\operatorname{loga}-\operatorname{logb}$ ……………. (i)
And hence, we need to determine the value of ‘a’ in terms of ‘b’.
Now, as we know the property of logarithm functions for subtraction of two $\log$ values can be given as following:
$\operatorname{logm}-\operatorname{logn}=\log \left( \dfrac{m}{n} \right)$ ……………. (ii)
Now, we can observe the left hand side of the equation (ii) and the right hand side of the equation (i) i.e. given in the problem. We notice that both are similar to each other and hence we can apply identity written in equation (ii) with the equation (i). so, we get
$\Rightarrow \log \left( a-b \right)=\operatorname{loga}-\operatorname{logb}=log\left( \dfrac{a}{b} \right)$
$\Rightarrow \log \left( a-b \right)=log\left( \dfrac{a}{b} \right)$ ……………… (iii)
Now, we know that logarithm of any number will be same if the numbers are same, and based upon this statement, we have one more property of logarithm function as:
If $\operatorname{logm}=\operatorname{logn}$ then
m = n ………………..(iv)
Now, we can observe the relation (iii) and hence get that $\log$ of values (a – b) and $\left( \dfrac{a}{b} \right)$ are same i.e. they are equal, it means values (a – b) and $\left( \dfrac{a}{b} \right)$ should be same as well from the equation (iv) as mentioned above. So, we can get equation (iii)
As, $\Rightarrow \log \left( a-b \right)=log\left( \dfrac{a}{b} \right)$
Hence, we get
$\dfrac{a-b}{1}=\dfrac{a}{b}$
Now, cross-multiply the above equation to get the value of a in terms of ‘b’ as
b (a – b) = a, b (a - b) = a
Now, multiply (a – b) by b in the left hand side of the equation. So, we get
$ab-{{b}^{2}}=a$
Now, subtract ‘ab’ to both sides of the equation, so, we get
$\begin{aligned} & ab-{{b}^{2}}-ab=a-ab, \\\ & ab-ab-{{b}^{2}}=a-ab, \\\ & -{{b}^{2}}=a-ab \\\ \end{aligned}$
Now, take ‘a’ as common factor from the L.H.S. of the equation; so, we get
$-{{b}^{2}}=a\left( 1-ab \right)$
Now, divide the whole expression by ‘1 – b’ to both sides of the equation. So, we get
$\begin{aligned} & \dfrac{-{{b}^{2}}}{\left( 1-b \right)}=\dfrac{a\left( 1-b \right)}{\left( 1-b \right)}, \\\ & \dfrac{-{{b}^{2}}}{\left( 1-b \right)}=a \\\ \end{aligned}$
Now, multiply and divide by ‘-1’ n the left hand side of the equation, we get
$\begin{aligned} & \dfrac{-1\left( -{{b}^{2}} \right)}{-1\left( 1-b \right)}=a, \\\ & \dfrac{{{b}^{2}}}{b-1}=a, \\\ & a=\dfrac{{{b}^{2}}}{b-1} \\\ \end{aligned}$
Hence, if $\log \left( a-b \right)=\log a-\log b$, the value of a in terms of $b\to \dfrac{{{b}^{2}}}{b-1}$.

Note: One may get confused with the formula of $\operatorname{loga}-\operatorname{logb}=log\left( \dfrac{a}{b} \right),\log a+\log b=\log ab$ and can replace them $\log a-\log b\Rightarrow \log ab$ in the problem, which is wrong. So, be clear with the properties of the logarithm functions.
One may notice that we did not take any base anywhere in the solution with the logarithm terms. As it is obvious that the bases of all of them are the same. So, we generally do not require to write bases with these kinds of questions. And we take base as ‘10’ if only $'\log '$ term is used without any base.