For the logarithm too, a base less than 11 implies a reversed direction (and the argument must remain >0>0). The five examples that follow cover the standard cases.

Example — log1/3(x+1)2\log_{1/3}(x+1)\le 2

EC: x+1>0    x>1x+1>0\iff x>-1. I rewrite 2=log1/3(1/9)2=\log_{1/3}(1/9): log1/3(x+1)log1/3(1/9).\log_{1/3}(x+1)\le\log_{1/3}(1/9). Base <1<1, direction reversed: x+11/9    x8/9x+1\ge 1/9 \iff x\ge -8/9. Intersection with the EC: x8/9\boxed{x\ge -8/9}.

Example — Decimal base 1/101/10

log1/10(2x3)>1\log_{1/10}(2x-3)>-1. EC: 2x3>0    x>3/22x-3>0\iff x>3/2. With 1=log1/10(10)-1=\log_{1/10}(10): log1/10(2x3)>log1/10(10)    2x3<10    x<13/2.\log_{1/10}(2x-3)>\log_{1/10}(10) \iff 2x-3<10 \iff x<13/2. Intersection: 3/2<x<13/2\boxed{3/2<x<13/2}.

Example — Substitution with base <1<1

(log1/2x)23log1/2x+20(\log_{1/2}x)^2 - 3\log_{1/2}x + 2 \ge 0. I set t=log1/2xt=\log_{1/2}x (with EC x>0x>0): t23t+20    t1t2.t^2-3t+2\ge 0 \iff t\le 1 \vee t\ge 2. I go back to xx:

  • log1/2x1    x1/2\log_{1/2}x\le 1 \iff x\ge 1/2 (direction reversed, base <1<1);
  • log1/2x2    x1/4\log_{1/2}x\ge 2 \iff x\le 1/4 (direction reversed).

Together with the EC (x>0x>0): 0<x1/4x1/2\boxed{0<x\le 1/4 \vee x\ge 1/2}.

Example — Comparison of two logs with base <1<1

log1/4(x2)>log1/4(3x7)\log_{1/4}(x-2)>\log_{1/4}(3x-7). EC: x2>0x-2>0 and 3x7>0    x>7/33x-7>0\iff x>7/3. Base <1<1, direction reversed: x2<3x7    5<2x    x>5/2x-2<3x-7 \iff 5<2x \iff x>5/2. Intersection with the EC: x>5/2\boxed{x>5/2}.

Example — Comparison with 00

log0,5(x24)>0\log_{0,5}(x^2-4)>0. EC: x24>0    x<2x>2x^2-4>0\iff x<-2\vee x>2. With 0=log0,510=\log_{0,5}1, base <1<1, direction reversed: x24<1    x2<5    5<x<5x^2-4<1 \iff x^2<5\iff -\sqrt{5}<x<\sqrt{5}. Intersection with the EC: 5<x<22<x<5\boxed{-\sqrt{5}<x<-2 \vee 2<x<\sqrt{5}}.

Topics: Logarithmic function
Concepts: Existence conditions · Logarithmic inequalities · Logarithm
Functions: Logarithmic function
Skills: Reason by cases · Solve inequalities