Question

If \[{\log _{0.2}}\left( {x - 1} \right) > {\log _{0.04}}\left( {x + 5} \right)\] then solve for the value of \[x\]
A. \[ - 1 < x < 4\]
B. \[2 < x < 3\]
C. \[1 < x < 4\]
D. \[1 < x < 3\]

Answer 1

Hint: We can cancel logarithms on both sides whenever they have the same bases and then the inequality sign is reversed. In a logarithmic function \[{\log _x}a\] is positive when \[a > 0\]. So, use this concept to reach the solution of the problem.

Complete step-by-step solution -
Given \[{\log _{0.2}}\left( {x - 1} \right) > {\log _{0.04}}\left( {x + 5} \right)\]
We can write \[0.04\]as \[{\left( {0.2} \right)^2}\] then we get
\[ \Rightarrow {\log _{0.2}}\left( {x - 1} \right) > {\log _{{{\left( {0.2} \right)}^2}}}\left( {x + 5} \right)\]
We know that \[{\log _{{x^2}}}a = \dfrac{1}{2}{\log _x}a\].
By using the above formula, we have
\[
   \Rightarrow {\log _{0.2}}\left( {x - 1} \right) > \dfrac{1}{2}{\log _{0.2}}\left( {x + 5} \right) \\
   \Rightarrow 2{\log _{0.2}}\left( {x - 1} \right) > {\log _{0.2}}\left( {x + 5} \right) \\
\]
Also, we know that \[2{\log _x}a = {\log _x}{a^2}\].
So, we have
\[ \Rightarrow {\log _{0.2}}{\left( {x - 1} \right)^2} > {\log _{0.2}}\left( {x + 5} \right)\]
Cancelling logarithms on both sides as they have common bases, then the inequality sign also gets reversed.
Hence, we have
\[
   \Rightarrow {\left( {x - 1} \right)^2} < \left( {x + 5} \right) \\
   \Rightarrow {x^2} - 2x + 1 < x + 5 \\
   \Rightarrow {x^2} - 2x - x + 1 - 5 < 0 \\
   \Rightarrow {x^2} - 3x - 4 < 0 \\
   \Rightarrow {x^2} - 4x + x - 4 < 0 \\
\]
Grouping the common terms, we have
\[
   \Rightarrow x\left( {x - 4} \right) + 1\left( {x - 4} \right) < 0 \\
   \Rightarrow \left( {x + 1} \right)\left( {x - 4} \right) < 0 \\
\]
We know that if \[\left( {x - a} \right)\left( {x - b} \right) < 0\] then it can be written as \[a < x < b\]
By using the above formula, we get
\[ \Rightarrow - 1 < x < 4\]
 Now consider \[{\log _{0.2}}\left( {x - 1} \right)\]
We know that in the logarithmic function \[{\log _x}a\] is positive when \[a > 0\]
So, in the logarithm \[{\log _{0.2}}\left( {x - 1} \right)\] we have \[x - 1 > 0\] i.e., \[x > 1\]
From \[x > 1\] and \[ - 1 < x < 4\] we get
\[\therefore 1 < x < 4\]
Thus, the correct option is C. \[1 < x < 4\]

Note: If \[\left( {x - a} \right)\left( {x - b} \right) < 0\] then it can be written as \[a < x < b\]. The equation \[\log x = 100\] is another way of writing \[{10^x} = 100\]. This relationship makes it possible to remove logarithms from an equation by raising both sides to the same exponent as the base of the logarithm.