Collection of Maths Problems

Inequalities with logarithms

Task number: 2764

• Variant 1

  \( \frac{2-\log x}{1+\log x}\ge 0 \)

• Resolution

  The expression is undefined for \(x\le 0\) and also when \(1+\log x=0\), i.e. \(x=\frac1{10}\).

  • \(0< x< \frac1{10}: 2-\log x\le 0 \longrightarrow x\ge 100 \longrightarrow \) no solution,
  • \(x > \frac1{10}: 2-\log x\ge 0 \longrightarrow x\le 100 \longrightarrow x\in \left(\frac{1}{10},100\right\rangle\)

• Result

  The solution is \(x\in \left(\frac{1}{10},100\right\rangle\).

• Variant 2

  \( \frac1{\log x}\ge \log x \)

• Resolution

  The expression is undefined when \(x\le 0\) and when \(\log x=0\) i.e. \(x=1\).

  • \(0< x< 1: 1\le \log^2 x \longrightarrow \)
    1. \(\log x \ge 1 \longrightarrow \) no solution
    2. \(\log x \le -1 \longrightarrow x\in \left(0, \frac1{10}\right\rangle\)
  • \(x>1: 1\ge \log^2 x \longrightarrow -1\le \log x \le 1 \longrightarrow \frac1{10} \le x \le 10 \longrightarrow x \in (1{,}10\rangle\)

• Result

  The solution is \(x\in \left(0, \frac1{10}\right\rangle \cup (1{,}10\rangle\).

• Variant 3

  \( \frac{\ln |x|}{4-x^2} \ge 0 \)

• Resolution

  The expression is undefined when \(|x|\le 0\) i.e. \(x=0\) and when \(4-x^2=0\) i.e. \(x=\pm 2\).

  • \(x< -2: \ln|x|\le 0 \longrightarrow \ln(-x)\le 0 \longrightarrow -x\le 1\longrightarrow x\ge -1 \longrightarrow\) no solution
  • \(-2\le x\le 0: \ln(-x)\ge 0 \longrightarrow x\le -1 \longrightarrow x\in (-2,-1\rangle\)
  • \(0\le x\le 2: \ln x\ge 0 \longrightarrow x\ge 1 \longrightarrow x\in \langle 1{,}2)\)
  • \(x>2: \ln x \le 0 \longrightarrow x\le 1 \longrightarrow\) no solution

• Result

  Ths solution is \(x\in (-2,-1\rangle \cup \langle 1{,}2)\).

• Variant 4

  \( \log_{\frac13}(x^2-3x+3)\ge 0 \)

• Hint

  To eliminate the base of the exponent use e.g. the identity \(\log_a b=\frac{\ln a}{\ln b}\).

• Resolution

  The expression is undefined when \(x^2-3x+3\le 0\), i.e. \( x_{1{,}2}=\frac{3\pm\sqrt{9-12}}{2} \) so the value of the polynomial \(x^2-3x+3\) is positive for all x.

  \( \frac{\ln (x^2-3x+3)}{\ln \frac13} \ge 0 \longrightarrow \ln (x^2-3x+3) \le 0 \longrightarrow x^2-3x+3 \le 1 \longrightarrow x^2-3x+2 \le 0 \longrightarrow \\ x_{1{,}2}=\frac{3\pm\sqrt{9-8}}{2} \longrightarrow x\in \langle 1{,}2\rangle \)

  In the first step we used the fact that \(\ln \frac13 <0\).

• Result

  The solution is \(x\in \langle 1{,}2\rangle\).