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Question: If we have a logarithmic inequality \[{{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.09}}\left( x-1 ...

If we have a logarithmic inequality log⁡0.3(x−1)<log⁡0.09(x−1){{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.09}}\left( x-1 \right), then find the interval in which x lies.
A). x > 2
B). x < 2
C). x > -2
D). none of these

Explanation

Solution

At first use the fact that in log⁡ab{{\log }_{a}}b, b is always positive. So, x–1>0x – 1 > 0 or x>1.x > 1. After that solve it further then use the fact that if log⁡ab<0{{\log }_{a}}b< 0 and a lies between 0 and 1 then we can write it as b>1b > 1 and hence get the value of x.

Complete step-by-step solution:
In the question we are given an equation log⁡0.3(x−1)<log⁡0.09(x−1){{\log }_{0.3}}\left( x-1 \right)<{{\log }_{0.09}}\left( x-1 \right) and we have to find values for x for which the given equation satisfies according to the given options.
So the equation in the question is,
log⁡0.3(x−1)<log⁡0.09(x−1){{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.09}}\left( x-1 \right)
In the term log⁡ab{{\log }_{a}}b the expression or term b should be positive according to the definition of logarithm.
So, x – 1 should be positive or greater than 0.
Hence, x–1>0x – 1 > 0 or x>1.x > 1.
Now we will use the identity that, log⁡a2b=12log⁡ab{{\log }_{{{a}^{2}}}}b=\dfrac{1}{2}{{\log }_{a}}b.
So, the given equation,
log⁡0.3(x−1)<log⁡0.09(x−1){{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.09}}\left( x-1 \right) can be written as,
log⁡0.3(x−1)<log⁡(0.3)2(x−1){{\log }_{0.3}}\left( x-1 \right)< {{\log }_{{{\left( 0.3 \right)}^{2}}}}\left( x-1 \right)
Or, log⁡0.3(x−1)<12log⁡0.3(x−1){{\log }_{0.3}}\left( x-1 \right)< \dfrac{1}{2}{{\log }_{0.3}}\left( x-1 \right).
Now multiplying by 2 on both the sides so we get,
2log⁡0.3(x−1)<log⁡0.3(x−1)2{{\log }_{0.3}}\left( x-1 \right)< {{\log }_{0.3}}\left( x-1 \right)
So, we can write it as,
log⁡0.3(x−1)<0{{\log }_{0.3}}\left( x-1 \right)< 0
Now if log⁡ab<0{{\log }_{a}}b< 0 and a is greater that 0 but less than 1 then we can write it as b > 1.
Now applying this we can write, log⁡0.3(x−1)<0{{\log }_{0.3}}\left( x-1 \right)< 0.
As, x–1>1x – 1 > 1
Or, x>2x > 2
Hence, the correct option is (a).

Note: Generally if an inequation is given let’s say log⁡ab<c{{\log }_{a}}b< c, where a, b, c are constants. If a is great equation than 1 then b<acb< {{a}^{c}}. This is the most common mistake students generally do while solving inequalities related to logarithms.