Logarithmic inequality

Statement

Solve the inequality: $\log_2(x+1) + \log_2(x-1) \le 3$

Solution

Recognising the strategy. Sum of logarithms in the same base: they are combined into a single logarithm, then monotonicity is exploited (base $2>1$, increasing function, direction preserved), without ever forgetting the existence conditions. Existence conditions: $x+1>0$ and $x-1>0$, hence $x>1$. Combining the logarithms: $\log_2\bigl[(x+1)(x-1)\bigr] \le 3$ Monotonicity (base $>1$, apply $t\mapsto 2^t$ preserving the direction): $(x+1)(x-1) \le 2^3 \;\Longrightarrow\; x^2 - 1 \le 8 \;\Longrightarrow\; x^2 \le 9 \;\Longrightarrow\; -3 \le x \le 3$ Intersection with the EC $x>1$: $\boxed{1 < x \le 3}$

Links

Topics: Unifying methods
Concepts: Existence conditions · Logarithmic inequality · Monotonic function
Functions: Logarithmic function
Skills: Reasoning by cases · Solving inequalities
Exercise type: Solving an inequality