Question

If $\log (a - b) = \log a - \log b$ then find the value of $a$ in terms of $b$ will be?

Answer 1

Hint:We have given a $\log $ function and we have to find the value of $'a'$ in term of $'b'$. For Firstly we have to solve the Right hand side of the question. On the right hand side we have$\log a - \log b$. We apply a property of $\log $on it. We get a single value of $\log $.

We equal this value with the left hand side and cancel $\log $with each other now. We will left with on equations in terms of $'a'$ and $'b'$ we solve this equation and find the value of a
term of $b$.

Complete step by step solution:
We have given a function
$\log (a - b) = \log a - \log b$
We have to find the value of $'a'$ so term of $'b'$.
Firstly we solve the Right hand side of the function.
$R.H.S. = \log a - \log b$
Now we have the property of $\log $function that
$\log m = \log n - \log \dfrac{m}{n}$
So we have $R.H.S. = \log a - \log b$
$ = \log \dfrac{m}{n}$
Now Left hand side is $\log (a - b)$
Equating Left hand side and Right hand side
$\log (a - b) = \log \dfrac{a}{b}$
This implies $a - b = \dfrac{a}{b}$
$a - \dfrac{a}{b} = b$
Taking \[a\] common from Left hand side
$a\left( {1 - \dfrac{1}{b}} \right) = b$
$a\left( {\dfrac{{b - 1}}{b}} \right) = b$
$b(b - 1) = {b^2}$
\[a = \dfrac{{{b^2}}}{{{b^{ - 1}}}}\]

So value of \[a = \dfrac{{{b^2}}}{{{b^{ - 1}}}}\]

Note: Logarithmic Function: A logarithmic function is inverse of exponential function. The logarithmic function is defined as for $x > 0,a > 0$ and \[a \ne 1,y = {\log _a}^x\]. If and only if . Then the function is given as $f(x) = {\log _a}^x$ .

The base of logarithm is $a$. This can be read as $\log $base $a$ of $x$. The most
common bases used in logarithmic function are base $10$and base$e$.