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Question: If \[f\left( x \right) = {\log _e}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)\],\[\left| x \right| < 1...

If f(x)=log⁡e(1−x1+x)f\left( x \right) = {\log _e}\left( {\dfrac{{1 - x}}{{1 + x}}} \right),∣x∣<1\left| x \right| < 1, thenf(2x1+x2)f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)is equal to:
A.2f(x)2f\left( x \right)
B.2f(x2)2f\left( {{x^2}} \right)
C.(f(x))2{\left( {f\left( x \right)} \right)^2}
D.(f(x))3{\left( {f\left( x \right)} \right)^3}

Explanation

Solution

Here we will find the value of f(2x1+x2)f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) by substituting the value of xx as 2x1+x2\dfrac{{2x}}{{1 + {x^2}}} in the function f(x)f\left( x \right). Then we will simplify the equation to get the answer in terms of the main function that is f(x)f\left( x \right).

Complete step-by-step answer:
Given function is f(x)=log⁡e(1−x1+x)f\left( x \right) = {\log _e}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)……………….(1)\left( 1 \right)
We will find the value of f(2x1+x2)f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right). Therefore, substituting the value of xx as 2x1+x2\dfrac{{2x}}{{1 + {x^2}}} in the equation(1)\left( 1 \right), we get
⇒f(2x1+x2)=log⁡e(1−2x1+x21+2x1+x2)\Rightarrow f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{1 - \dfrac{{2x}}{{1 + {x^2}}}}}{{1 + \dfrac{{2x}}{{1 + {x^2}}}}}} \right)
Now, we will solve and simplify the above equation. So, by taking 1+x21 + {x^2} common in both the numerator and denominator, we get
⇒f(2x1+x2)=log⁡e(1+x2−2x1+x21+x2+2x1+x2)=log⁡e(1+x2−2x1+x2+2x)\Rightarrow f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{\dfrac{{1 + {x^2} - 2x}}{{1 + {x^2}}}}}{{\dfrac{{1 + {x^2} + 2x}}{{1 + {x^2}}}}}} \right) = {\log _e}\left( {\dfrac{{1 + {x^2} - 2x}}{{1 + {x^2} + 2x}}} \right)
Now, we can clearly see that the numerator and the denominator is the perfect square of 1−x1 - x and 1+x1 + x respectively. So, we get
⇒f(2x1+x2)=log⁡e((1−x)2(1+x)2)=log⁡e(1−x1+x)2\Rightarrow f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = {\log _e}\left( {\dfrac{{{{\left( {1 - x} \right)}^2}}}{{{{\left( {1 + x} \right)}^2}}}} \right) = {\log _e}{\left( {\dfrac{{1 - x}}{{1 + x}}} \right)^2}
Now as we know this the property of the logarithmic function that log⁡ab=blog⁡a\log {a^b} = b\log a.
Applying the property of logarithmic function, we get
⇒f(2x1+x2)=2log⁡e(1−x1+x)\Rightarrow f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = 2{\log _e}\left( {\dfrac{{1 - x}}{{1 + x}}} \right)
Now from the equation (1)\left( 1 \right) we know that log⁡e(1−x1+x){\log _e}\left( {\dfrac{{1 - x}}{{1 + x}}} \right) is equal to f(x)f\left( x \right). Therefore, we can write
⇒f(2x1+x2)=2f(x)\Rightarrow f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) = 2f\left( x \right)
Hence, f(2x1+x2)f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) is equal to 2f(x)2f\left( x \right).
So, option A is the correct option.

Note: Here, it is important to rewrite the function whose value is to be found out in such a way that the function changes in terms of the given value. So, that we can easily substitute the values and find the answer. Also to solve this question we need to keep in mind the basic logarithmic properties. Few properties of the logarithmic function is:
(1)log⁡a+log⁡b=log⁡ab\log a + \log b = \log ab
(2)log⁡a−log⁡b=log⁡ab\log a - \log b = \log \dfrac{a}{b}
(3)log⁡ab=blog⁡a\log {a^b} = b\log a