How do we solve $\log (x - 3) + \log (x - 5) = \log (2x - 9)$ ?

Hint: To solve this question, first we will use the logarithm Product Rule and then rewrite the expression in the exponential form to avoid the log or to remove the log from the both sides. After removal of log, we can easily get the factors of the equation. That’s how we can solve this question.

Complete step by step answer:
Given logarithmic expression:-
$\log (x - 3) + \log (x - 5) = \log (2x - 9)$
To solve this expression, we have to go through steps as follows:-
Step-1: Use Logarithm Product Rule, i.e..
\[\because \log a + \log b = \log (a.b)\]
Now,
\[ \Rightarrow \log [(x - 3)(x - 5)] = \log (2x - 9)\]
$ \Rightarrow \log ({x^2} - 3x - 5x + 15) = \log (2x - 9)$
Step-2: Rewrite in exponential form with base to (“drop” log since we have same log both side of expression)
\[
\Rightarrow {10^{\log ({x^2} - 8x + 15)}} = {10^{\log (2x - 9)}} \\
   \Rightarrow {x^2} - 8x + 15 = 2x - 9 \\
\]
Step-3: Now, write the above quadratic equation in its standard form, $a{x^2} + bx + c = 0$ :
$
\Rightarrow {x^2} - 8x + 15 = 2x - 9 \\
  \because {x^2} - 10x + 24 = 0 \\
\ $
Step-4: Now, the above equation is solved by factorization:
$ \Rightarrow (x - 12)(x + 2) = 0$
$\therefore \,\,x - 12 = 0$
$ \Rightarrow x = 12$
And,
$
\Rightarrow x + 2 = 0 \\
   \Rightarrow x = - 2 \\
 $
Hence, in the given equation, the value of $x$ is 12.

Note: The argument for the logarithm cannot be a negative number. That’s why we couldn’t take -2 as the value of $x$ for the given expression:
Check $x = - 2$ :
\[
\Rightarrow \log ( - 2 - 3) + \log ( - 2 - 5) = \log (2. - 2 - 9) \\
   \Rightarrow \log ( - 5) + \log ( - 7) = \log ( - 13) \\
\]
So, we can’t solve it further.