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. $1 < x < 3$

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.