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Question: If we have a logarithmic inequality as \({{\log }_{0.5}}(x-1)<{{\log }_{0.25}}(x-1)\) , then x lies ...

If we have a logarithmic inequality as log⁡0.5(x−1)<log⁡0.25(x−1){{\log }_{0.5}}(x-1)<{{\log }_{0.25}}(x-1) , then x lies in the interval,
a) (2,∞\infty )
b) (3, ∞\infty )
c) (- ∞\infty , 0)
d) (0,3)

Explanation

Solution

We will solve the RHS of the equation. We will make the base of the logarithm same on both the side. As the base of the logarithm are the same so the log can be removed. We will know to solve the inequality and we will get the interval of x.

Complete step-by-step solution:
We have equation log⁡0.5(x−1)<log⁡0.25(x−1){{\log }_{0.5}}(x-1)<{{\log }_{0.25}}(x-1),
The RHS side of the equation can be written as,
⇒log⁡0.5(x−1)<log⁡(0.5)2(x−1)\Rightarrow {{\log }_{0.5}}(x-1)<{{\log }_{{{\left( 0.5 \right)}^{2}}}}(x-1)
We know that
⇒log⁡a2b=12log⁡ab\Rightarrow {{\log }_{{{a}^{2}}}}b=\dfrac{1}{2}{{\log }_{a}}b
So RHS can be written as,
⇒log⁡0.5(x−1)<12log⁡0.5(x−1) ⇒2log⁡0.5(x−1)<log⁡0.5(x−1) \begin{aligned} & \Rightarrow {{\log }_{0.5}}(x-1)<\dfrac{1}{2}{{\log }_{0.5}}(x-1) \\\ & \Rightarrow 2{{\log }_{0.5}}(x-1)<{{\log }_{0.5}}(x-1) \\\ \end{aligned}
We know that
⇒2log⁡ab=log⁡ab2\Rightarrow 2{{\log }_{a}}b={{\log }_{a}}{{b}^{2}}
So, the above equation can be written as,
⇒log⁡0.5(x−1)2<log⁡0.5(x−1)\Rightarrow {{\log }_{0.5}}{{(x-1)}^{2}}<{{\log }_{0.5}}(x-1)
We know that if log⁡ab<log⁡ac{{\log }_{a}}b<{{\log }_{a}}c and is 0c
⇒(x−1)2>(x−1)\Rightarrow {{(x-1)}^{2}}>(x-1)
We will now solve the above inequality
⇒(x−1)2>(x−1) ⇒x2+1−2x>x−1 ⇒x2−3x+2>0 ⇒x2−2x−x+2>0 ⇒x(x−2)−1(x−2) ⇒(x−1)(x−2)>0 \begin{aligned} & \Rightarrow {{(x-1)}^{2}}>(x-1) \\\ & \Rightarrow {{x}^{2}}+1-2x>x-1 \\\ & \Rightarrow {{x}^{2}}-3x+2>0 \\\ & \Rightarrow {{x}^{2}}-2x-x+2>0 \\\ & \Rightarrow x(x-2)-1(x-2) \\\ & \Rightarrow \left( x-1 \right)\left( x-2 \right)>0 \\\ \end{aligned}
If we put the value of x < 1, we will get a positive value and if we put the value of x > 2 we will get a positive value. If we put a value of 12 the answer is positive.
So, interval of x will be (−∞,1)⋃(2,∞)\left( -\infty ,1 \right)\bigcup \left( 2,\infty \right)
But we know that in log⁡ab{{\log }_{a}}b b>0 so if we put x−1>0x-1 >0 we get x>1x> 1, so the interval of x will be (2,∞)\left( 2,\infty \right)
So the answer is a)(2,∞)\left( 2,\infty \right)

Note: Before attending this question, students must have a thorough knowledge of logarithmic identities. The most important thing in this question is to change the sign of inequality when we remove the log from both the side, students tend to make mistakes in this area. It is important to find all the possible interval for x by representing the answers that we get number line and then we can eliminate the unnecessary intervals as per the question.