Answer :
Let's examine the steps given in the problem and identify the mistake in solving the inequality.
Initial Problem:
[tex]\log_{0.2}\left(\frac{x+2}{x}\right) \le 1[/tex]Change of Base and Rewriting the Logarithm:
The student attempts to rewrite the expression using a change of base:
[tex]\log_{0.2} A = \frac{\log_5 A}{\log_5 0.2}[/tex]
where [tex]A = \frac{x+2}{x}[/tex].Finding the Mistake:
In the provided solution steps, instead of correctly changing the base of the logarithm, the student incorrectly applies a negation and changes the expression:- The original inequality was [tex]\log_{0.2}(\frac{x+2}{x}) \le 1[/tex],
- It should have been converted to the same base without inversion or negation immediately: [tex]\log_5(\frac{x+2}{x}) \ge \log_5(0.2)[/tex].
- The error starts where the log was negated and rewritten as: [tex]\log_5(\frac{x+2}{x})^{-1} \le \log_5 5[/tex].
Correct Approach:
Begin correctly changing the inequality according to logarithmic properties:
Since [tex]0.2 = 5^{-1}[/tex],
[tex]\log_{0.2}\left(\frac{x+2}{x}\right) = -\log_5\left(\frac{x+2}{x}\right)[/tex]
Thus,
[tex]-\log_5\left(\frac{x+2}{x}\right) \le 1[/tex]
Multiplying both sides of the inequality by -1 (which reverses the inequality):
[tex]\log_5\left(\frac{x+2}{x}\right) \ge -1[/tex]
Therefore:
[tex]\frac{x+2}{x} \ge 5^{-1} = \frac{1}{5}[/tex]Solving the Correct Inequality:
[tex]\frac{x+2}{x} \ge \frac{1}{5}[/tex]
Multiply both sides by [tex]5x[/tex] to clear the fraction:
[tex]5(x+2) \ge x[/tex]
[tex]5x + 10 \ge x[/tex]
Rearrange:
[tex]4x \ge -10[/tex]
Divide by 4:
[tex]x \ge -\frac{5}{2}[/tex]
Thus, the correct solution range for [tex]x[/tex] is [tex]x \ge -\frac{5}{2}[/tex]. The mistake was in the handling of logarithmic properties and incorrectly applying operations to the inequality.