Question

Given that \[{{\log }_{0.5}}(x-1)<{{\log }_{0.25}}(x-1)\]. Find the interval in which x lies.

Answer 1

Hint: Now we will use properties of log to solve the problem first we can clearly see that 0.25 is square of 0.5 and we know ${{\log }_{{{a}^{n}}}}b=\dfrac{1}{n}{{\log }_{a}}b$ and also $n{{\log }_{a}}b={{\log }_{a}}{{b}^{n}}$ using this two we will try to write an equation with common base. Hence we will finally get an inequality in x and solve it to get the condition on x

Complete step by step answer:
Now the given equation is \[{{\log }_{0.5}}(x-1)<{{\log }_{0.25}}(x-1)\].
Now since 0.25 is square of 0.5 we can write the above equation as
${{\log }_{(0.5)}}(x-1)<{{\log }_{{{(0.5)}^{2}}}}(x-1)$
Now we know ${{\log }_{{{a}^{n}}}}b=\dfrac{1}{n}{{\log }_{a}}b$ using this property we get the equation as
${{\log }_{(0.5)}}(x-1)<\dfrac{1}{2}{{\log }_{0.5}}(x-1)$
Taking 2 on RHS we get
$2{{\log }_{(0.5)}}(x-1)<{{\log }_{0.5}}(x-1)$
Now we have another property of log which says $n{{\log }_{a}}b={{\log }_{a}}{{b}^{n}}$
Using this property in the above equation we get.
${{\log }_{0.5}}{{(x-1)}^{2}}<{{\log }_{0.5}}(x-1)$
Now we know that ${{\log }_{a}}M<{{\log }_{a}}N\Rightarrow M < N\text{ if }a > 1\text{ and }M > N\text{ if }a < 1$
Let us check with examples to understand this.
Now consider log with base 10 > 1.
$\begin{align}
  & {{\log }_{10}}10=1 \\
 & {{\log }_{10}}100={{\log }_{10}}{{10}^{2}}=2{{\log }_{10}}10=2 \\
 & \Rightarrow {{\log }_{10}}10<{{\log }_{10}}100 \\
\end{align}$
And also now we will consider log with base $\dfrac{1}{10}$ < 1
 \[\begin{align}
  & {{\log }_{\dfrac{1}{10}}}10={{\log }_{{{10}^{-1}}}}10=-{{\log }_{10}}10=-1 \\
 & {{\log }_{\dfrac{1}{10}}}100={{\log }_{{{10}^{-1}}}}{{10}^{2}}=-2{{\log }_{10}}10=-2 \\
 & \Rightarrow {{\log }_{\dfrac{1}{10}}}100<{{\log }_{\dfrac{1}{10}}}10 \\
\end{align}\]
Hence we can say that ${{\log }_{a}}M<{{\log }_{a}}N\Rightarrow M < N\text{ if }a > 1\text{ and }M>N\text{ if }a < 1$
Here 0.5 < 1. Hence the above equation becomes
$\begin{align}
  & {{(x-1)}^{2}}>(x-1) \\
 & \Rightarrow {{(x-1)}^{2}}-(x-1)>0 \\
 & \Rightarrow (x-1)[(x-1)-(1)]>0 \\
 & \Rightarrow (x-1)(x-2)>0............................(1) \\
\end{align}$
Now we want to solve an inequality (x-1)(x-2) > 0
Consider the expression (x-1).(x-2)
Here x=1 and x=2 are critical points at which the value is zero
For x lying between 1 and 2 let us say 1.5 we get (1.5-1) × (1.5-2) = (0.5) × (-0.5)=-0.25 < 0
Hence (x-1) × (x-2) < 0 between x=1 and x=2 ……….. (2)
Now for x > 2. Let us take x = 3
Then (3-1) × (3-2) = (2) × (1) = 2 > 0
Hence (x-1) × (x-2) > 0 for x > 2…………………….. (3)
Also for x < 1
Let us take x = 0
Then we get (0-1) × (0-2) = (-1) × (-2) = 2
Hence (x-1) × (x-2) > 0 for x < 1………….. (4)
Hence from equation (2), equation (3) and equation (4) we get solution of (x-1) (x-2) > 0 is
x < 1 or x > 2
Now note that for ${{\log }_{0.25}}(x-1)$ to be defined (x-1) > 0.
Hence x should always be greater than 1.
i.e. x > 1 …………..(5)
From equation (4) and equation (5) we get the solution of the equation as (x > 2)
Hence x lies between $(2,\infty )$
Hence x belongs to interval $(2,\infty )$

So, the correct answer is “Option A”.

Note: Now note that the sign of inequality changes if the base of log is less than 1 because of property ${{\log }_{a}}M<{{\log }_{a}}N\Rightarrow M < N\text{ if a > 1 and M > N if a < 1}$. Hence if a < 1 we take reverse the equality. Also note that $n{{\log }_{a}}b={{\log }_{a}}{{b}^{n}}$ while ${{\log }_{{{a}^{n}}}}b=\dfrac{1}{n}{{\log }_{a}}b$ this two properties are similar and confusing so should be used with caution.