Question

If \[{\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)\] then x lies in the interval
A. \[( - \infty ,1)\]
B. \[(1,2)\]
C. \[(2,\infty )\]
D. None of these

Answer 1

Hint: It's basically a problem based on Logarithmic properties we will mainly use two properties here, one is \[{\log _{{a^n}}}b = \dfrac{1}{n}{\log _a}b\] and the other one is \[{\log _a}b = \dfrac{{{{\log }_n}b}}{{{{\log }_n}a}}\] where n is any number we can choose as base of log.

Complete step by step answer:
So here we are given that \[{\log _{0.3}}(x - 1) < {\log _{0.09}}(x - 1)\]
So let's get everything in one side, it will look like as
\[{\log _{0.3}}(x - 1) - {\log _{0.09}}(x - 1) < 0\]
Now we know that \[0.09 = \dfrac{9}{{100}} = {\left( {\dfrac{3}{{10}}} \right)^2}\]
So we can write
\[{\log _{0.3}}(x - 1) - {\log _{{{\left( {0.3} \right)}^2}}}(x - 1) < 0\]
So now If we use the property \[{\log _{{a^n}}}b = \dfrac{1}{n}{\log _a}b\]
We will get the whole thing as
\[\begin{array}{l}
 \Rightarrow {\log _{0.3}}(x - 1) - \dfrac{1}{2}{\log _{0.3}}(x - 1) < 0\\
 \Rightarrow \dfrac{1}{2}{\log _{0.3}}(x - 1) < 0
\end{array}\]
So now it can be easily seen that
\[ \Rightarrow {\log _{0.3}}(x - 1) < 0\]
Now again using the property \[{\log _a}b = \dfrac{{{{\log }_n}b}}{{{{\log }_n}a}}\] we will get this as
\[ \Rightarrow \dfrac{{{{\log }_{10}}(x - 1)}}{{{{\log }_{10}}(0.3)}} < 0\]
Now in this case it must be clear that \[ \Rightarrow \dfrac{{{{\log }_{10}}(x - 1)}}{{{{\log }_{10}}(0.3)}} < 0\] is holding until and unless \[{{{\log }_{10}}(0.3)}\] is in the denominator because it’s value is negative so if we multiply it with the 0 on RHS the sign will change, which means that, we will be left with
\[\begin{array}{l}
 \Rightarrow {\log _{10}}(x - 1) > 0\\
 \Rightarrow x - 1 > {10^0}\\
 \Rightarrow x - 1 > 1\\
 \Rightarrow x > 2
\end{array}\]
So from here we can say that we will be left with All values greater than 2
Which is \[(2,\infty )\]

So, the correct answer is “Option C”.

Note: The key to solve this question are the properties of log used in it there are some other important properties of log as well, such as
\[\begin{array}{l}
{\log _b}(MN) = {\log _b}M + {\log _b}N\\
{\log _b}\left( {\dfrac{M}{N}} \right) = {\log _b}M - {\log _b}N\\
{\log _b}\left( {{M^p}} \right) = p{\log _b}M
\end{array}\]