Answer

Verified

61.2k+ views

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 $x$.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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.

Recently Updated Pages

Write a composition in approximately 450 500 words class 10 english JEE_Main

Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main

What is the common property of the oxides CONO and class 10 chemistry JEE_Main

What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main

If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main

The area of square inscribed in a circle of diameter class 10 maths JEE_Main

Other Pages

A square frame of side 10 cm and a long straight wire class 12 physics JEE_Main

Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main

A pilot in a plane wants to go 500km towards the north class 11 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main