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Question: If we have a logarithmic inequality \({\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)\), then \(x\) lie...

If we have a logarithmic inequality log⁡0.3(x−1)<log⁡0.09(x−1){\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1), then xx lies in the interval
A)(2,∞)A)(2,\infty )
B)(1,2)B)(1,2)
C)(−2,−1)C)( - 2, - 1)
D)D) None of these

Explanation

Solution

First, we will understand what the logarithmic operator represents in mathematics. A logarithm function or log operator is used when we have to deal with the powers and base of a number, to understand it better which is log⁡xm=mlog⁡x\log {x^m} = m\log x
It’s basically a problem based on the logarithmic properties. We will mainly use two properties of the logarithm to solve this problem.
Formula used:

Using the logarithm law, log⁡yx=log⁡xlog⁡y{\log _y}x = \dfrac{{\log x}}{{\log y}}
log⁡anb=1nlog⁡ab{\log _{{a^n}}}b = \dfrac{1}{n}{\log _a}b where nn is any number that we can choose as the base of the log.

Complete step-by-step solution:
Since from given that we have a relation inequality log⁡0.3(x−1)<log⁡0.09(x−1){\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)
Let just make it into an equation form, with all the values at one side then we have log⁡0.3(x−1)−log⁡0.09(x−1)<0{\log _{0.3}}(x - 1) - {\log _{0.09}}(x - 1) < 0
Since we know that 0.090.09 represented into the fraction form of 0.09=91000.09 = \dfrac{9}{{100}} and this can be rewritten using the square root as 0.09=9100=(310)20.09 = \dfrac{9}{{100}} = {(\dfrac{3}{{10}})^2}
Thus, we get the values as log⁡0.3(x−1)−log⁡0.09(x−1)<0⇒log⁡0.3(x−1)−log⁡(0.3)2(x−1)<0{\log _{0.3}}(x - 1) - {\log _{0.09}}(x - 1) < 0 \Rightarrow {\log _{0.3}}(x - 1) - {\log _{{{(0.3)}^2}}}(x - 1) < 0 with the base value of 1010
Now we apply the logarithm property that log⁡anb=1nlog⁡ab{\log _{{a^n}}}b = \dfrac{1}{n}{\log _a}b in the above equation, then we get log⁡0.3(x−1)−log⁡(0.3)2(x−1)<0⇒log⁡0.3(x−1)−12log⁡0.3(x−1)<0{\log _{0.3}}(x - 1) - {\log _{{{(0.3)}^2}}}(x - 1) < 0 \Rightarrow {\log _{0.3}}(x - 1) - \dfrac{1}{2}{\log _{0.3}}(x - 1) < 0 where n=2n = 2 and the base is 1010
Since we know that 1−12=121 - \dfrac{1}{2} = \dfrac{1}{2} and in the above equation both the values are the same and we apply this rule then we get log⁡0.3(x−1)−12log⁡0.3(x−1)<0⇒12log⁡0.3(x−1)<0{\log _{0.3}}(x - 1) - \dfrac{1}{2}{\log _{0.3}}(x - 1) < 0 \Rightarrow \dfrac{1}{2}{\log _{0.3}}(x - 1) < 0
Multiplied both sides with 22then we have log⁡0.3(x−1)<0{\log _{0.3}}(x - 1) < 0
Again applying the log property of log⁡yx=log⁡xlog⁡y{\log _y}x = \dfrac{{\log x}}{{\log y}}, then we get log⁡0.3(x−1)<0⇒log⁡10(x−1)log⁡10(0.3)<0{\log _{0.3}}(x - 1) < 0 \Rightarrow \dfrac{{{{\log }_{10}}(x - 1)}}{{{{\log }_{10}}(0.3)}} < 0 is holding until and unless log⁡10(0.3){\log _{10}}(0.3) is in the denominator because its value is negative so if we multiply it with the zero on the right side then the sign will change, which means that we will be left with log⁡10(x−1)>0{\log _{10}}(x - 1) > 0
Hence, we solve further, we get log⁡10(x−1)>0⇒x−1>100⇒x−1>1⇒x>2{\log _{10}}(x - 1) > 0 \Rightarrow x - 1 > {10^0} \Rightarrow x - 1 > 1 \Rightarrow x > 2
So, from here we say that we will be left with a value greater than 22 for the x, and it may go up to any undetermined value.
Hence, we get x∈(2,∞)x \in (2,\infty ) and thus the interval lies in (2,∞)(2,\infty )
Therefore, the option A)(2,∞)A)(2,\infty ) is correct.

Note: The key to solving this problem is the logarithm functions, by using its properties we solved. Some other important logarithm properties are log⁡(ab)=log⁡a+log⁡b\log (ab) = \log a + \log b, log⁡(ab)=log⁡a−log⁡b\log (\dfrac{a}{b}) = \log a - \log b. The intervals are measured by using the greater than sign, which is x>2x > 2 so the value of x must be at least 22 and at most, it can be anything which means undetermined infinity.