Question

Solve the following equation:
\[\log (x + 1) + \log (x - 1) = \log 24\]

Answer 1

Hint: We are given to solve a logarithmic equation. We will first simplify it by using the sum of logarithmic functions. In this way we will be able to remove log function from the given equation and convert it into a simple quadratic equation. Then on solve the quadratic equation, we will get the value of variable \[x\]
Formula used: Summation of two log functions \[\log a\] and \[\log b\] is given by
\[\log a + \log b = \log ab\]
If we have \[\log a = \log b\], then we say that,
\[\log a = \log b \Leftrightarrow a = b\]
We have also used the following identity here.
\[(a + b)(a - b) = {a^2} - {b^2}\]

Complete step-by-step solution:
We are given to solve the equation,
\[\log (x + 1) + \log (x - 1) = \log 24\]
We know that, the formula for summation of any two log functions \[\log a\] and \[\log b\] is given below as,
\[\log a + \log b = \log ab\]
Using, the above formula, we simplify the given equation as,
\[ \Rightarrow \log [(x + 1)(x - 1)] = \log 24\],
Now, on using the formula \[(a + b)(a - b) = {a^2} - {b^2}\] in the above equation, we get,
\[ \Rightarrow \log ({x^2} - 1) = \log 24\]
We know that \[\log a = \log b \Leftrightarrow a = b\], then,
\[
   \Rightarrow ({x^2} - 1) = 24 \\
   \Rightarrow {x^2} = 25 \\
   \Rightarrow x = \pm 5 \]
Since log cannot take negative values, the value of \[x\] cannot be \[ - 5\], so we can only take the value of \[x\] as \[5\].
So, our solution for the given equation will be, \[x = 5\].

Note: This is to note that the domain of logarithmic function is the set of all positive numbers. Hence, it cannot take the zero and negative values. This is why we have neglected \[ - 5\] as our answer. Although, the range of logarithmic functions is set of all real numbers. Hence, it can give all the real numbers as an answer.
Additional information: Logarithmic functions are the inverse of the exponential functions and can be converted into them.