Question

# If $\log \left( {1 + x} \right) + \log \left( {x - 1} \right) = \log 8$ , then $x$ is equal to${\text{A}}{\text{. 2}} \\ {\text{B}}{\text{. 3}} \\ {\text{C}}{\text{. - 3}} \\ {\text{D}}{\text{. - 2}} \\$

Hint: In this question we have to find the value of $x$, so the key concept is to apply the basic logarithmic identities in the given equation $\log \left( {1 + x} \right) + \log \left( {x - 1} \right) = \log 8$ to get the correct value of $x$.

We have been given that $\log \left( {1 + x} \right) + \log \left( {x - 1} \right) = \log 8$ …………. (1)
We know that, if $a > 0,b > 0$ then we have,
$\Rightarrow \log a + \log b = \log ab$
So, equation (1) can also be written as
$\Rightarrow \log (1 + x) + \log (x - 1) = \log 8 \\ \Rightarrow \log \left\{ {\left( {1 + x} \right)(x - 1)} \right\} = \log 8 \\$ ………… (2)
Now we can write $\left( {1 + x} \right)\left( {x - 1} \right) = {x^2} - {1^2} = {x^2} - 1$
So, equation (2) will become
$\Rightarrow \log ({x^2} - 1) = \log 8$ ………….. (3)
We also know that if $\log a = \log b$ then we have,
$\Rightarrow a = b$ for all $a > 0,b > 0$
So, equation (3) will become
$\Rightarrow \log ({x^2} - 1) = \log 8 \\ \Rightarrow {x^2} - 1 = 8 \\ \Rightarrow {x^2} = 9 \\ \Rightarrow x = \pm \sqrt 9 \\ \Rightarrow x = \pm 3 \\ \Rightarrow x = + 3, - 3 \\$
Here we get the two values of $x = + 3, - 3$
And we have the equation (1) is $\log \left( {1 + x} \right) + \log \left( {x - 1} \right) = \log 8$
So, for $x = - 3$ equation (1) is not defined because ‘$\log$’ is only defined for positive real numbers.
But for $x = + 3$ equation (1) is defined.
Hence option B is the correct answer.

Note: Whenever we face such types of problems the key point is that to simplify the problems by using basic logarithmic identities. So we have always remembered the basic logarithmic identities. After getting the solutions the most important step is rechecking whether the given equation will define or not for our founded solutions. The solution for which the given equation is defined is our right answer.