A graded collection of 20 practice problems on logarithmic inequalities, designed to build a firm understanding of domain conditions, monotonicity of the logarithmic function, direction reversal, and the final intersection with the domain.
Exercise 1 — level ★☆☆☆☆
Solve:
\[\log_2(x-1)>3\]
Answer
\[(9,+\infty)\]
Solution
Domain. The logarithm is defined only when its argument is positive:
\(x - 1 > 0 \quad \Longrightarrow \quad x > 1\).
So the domain is \(\mathcal{D} = (1, +\infty)\).
The base is \(2 > 1\), so the logarithmic function is increasing. The direction of the inequality is preserved:
\(\log_2(x-1) > 3 \quad \Longleftrightarrow \quad x-1 > 2^3 = 8 \quad \Longrightarrow \quad x > 9\).
Intersecting with the domain: \(x > 9\) is already contained in \(x > 1\), so the solution set is \((9, +\infty)\).
Exercise 2 — level ★☆☆☆☆
Solve:
\[\log_3(2x+1)\leq 2\]
Answer
\[\left(-\frac{1}{2},4\right]\]
Solution
Domain. The argument must be positive:
\(2x + 1 > 0 \quad \Longrightarrow \quad x > -\frac{1}{2}\).
So \(\mathcal{D} = \left(-\frac{1}{2}, +\infty\right)\).
Base \(3 > 1\), so the function is increasing and the direction is preserved:
\(\log_3(2x+1) \leq 2 \quad \Longleftrightarrow \quad 2x+1 \leq 3^2 = 9 \quad \Longrightarrow \quad 2x \leq 8 \quad \Longrightarrow \quad x \leq 4\).
Intersecting with the domain: \(x > -\frac{1}{2}\) and \(x \leq 4\), giving \(\left(-\frac{1}{2}, 4\right]\).
Exercise 3 — level ★☆☆☆☆
Solve:
\[\log_{1/2}(x+4)>-1\]
Answer
\[(-4,-2)\]
Solution
Domain. \(x + 4 > 0 \quad \Longrightarrow \quad x > -4\).
So \(\mathcal{D} = (-4, +\infty)\).
Base \(\frac{1}{2} < 1\), so the function is decreasing and the direction of the inequality is reversed:
\(\log_{1/2}(x+4) > -1 \quad \Longleftrightarrow \quad x+4 < \left(\frac{1}{2}\right)^{-1} = 2 \quad \Longrightarrow \quad x < -2\).
Intersecting with the domain: \(-4 < x < -2\), giving \((-4, -2)\).
Exercise 4 — level ★☆☆☆☆
Solve:
\[\log_{1/3}(5-x)\leq 1\]
Answer
\[\left(-\infty,\frac{14}{3}\right]\]
Solution
Domain. \(5 - x > 0 \quad \Longrightarrow \quad x < 5\).
Base \(\frac{1}{3} < 1\), so the function is decreasing and the direction is reversed:
\(\log_{1/3}(5-x) \leq 1 \quad \Longleftrightarrow \quad 5-x \geq \frac{1}{3}\).
\(-x \geq \frac{1}{3} - 5 = -\frac{14}{3}\). Multiplying by \(-1\) (which reverses the direction): \(x \leq \frac{14}{3}\).
Intersecting with the domain: \(\left(-\infty, \frac{14}{3}\right]\).
Exercise 5 — level ★★☆☆☆
Solve:
\[\log_2(x-1)+\log_2(x-3)>3\]
Answer
\[(5,+\infty)\]
Solution
Domain. Both arguments must be positive: \(x-1>0\) and \(x-3>0\), so \(x>3\).
Logarithm product rule: \(\log_2[(x-1)(x-3)] > 3\).
Base \(2>1\): \((x-1)(x-3) > 8 \iff x^2 - 4x - 5 > 0 \iff (x-5)(x+1) > 0\).
Solutions: \(x < -1\) or \(x > 5\).
Intersecting with the domain (\(x>3\)): \((5, +\infty)\).
Exercise 6 — level ★★☆☆☆
Solve:
\[\log_3(x+2)-\log_3(x-1)<1\]
Answer
\[\left(\frac{5}{2},+\infty\right)\]
Solution
Domain. \(x+2>0\) and \(x-1>0\), so \(x>1\).
Logarithm quotient rule: \(\log_3\left(\frac{x+2}{x-1}\right) < 1\).
Base \(3>1\): \(\frac{x+2}{x-1} < 3\).
Since \(x-1>0\) in the domain, we may multiply both sides without reversing the direction: \(x+2 < 3(x-1) \iff x > \frac{5}{2}\).
Intersecting with the domain: \(\left(\frac{5}{2}, +\infty\right)\).
Exercise 7 — level ★★☆☆☆
Solve:
\[\log_5(2x+1)\geq \log_5(x+4)\]
Answer
\[[3,+\infty)\]
Solution
Domain. \(x > -\frac{1}{2}\).
Base \(5>1\), increasing function: \(2x+1 \geq x+4 \iff x \geq 3\).
Intersecting with the domain: \([3, +\infty)\).
Exercise 8 — level ★★☆☆☆
Solve:
\[\log_{1/2}(x+5) < \log_{1/2}(3x-1)\]
Answer
\[\left(\frac{1}{3},3\right)\]
Solution
Domain. \(x+5>0\) and \(3x-1>0\), so \(x > \frac{1}{3}\).
Base \(\frac{1}{2} < 1\), decreasing function: direction reversed.
\(x+5 > 3x-1 \iff 6 > 2x \iff x < 3\).
Intersecting with the domain: \(\left(\frac{1}{3}, 3\right)\).
Exercise 9 — level ★★☆☆☆
Solve:
\[\log_2 x + \log_4 x \geq 6\]
Answer
\[[16,+\infty)\]
Solution
Domain. \(x > 0\).
\(\log_4 x = \frac{\log_2 x}{2}\). Let \(t = \log_2 x\):
\(t + \frac{t}{2} \geq 6 \iff \frac{3t}{2} \geq 6 \iff t \geq 4 \iff x \geq 16\).
Intersecting with the domain: \([16, +\infty)\).
Exercise 10 — level ★★☆☆☆
Solve:
\[\log_3 x - \log_9 x < 2\]
Answer
\[(0,81)\]
Solution
Domain. \(x > 0\).
\(\log_9 x = \frac{\log_3 x}{2}\). Let \(t = \log_3 x\):
\(t - \frac{t}{2} < 2 \iff \frac{t}{2} < 2 \iff t < 4 \iff x < 81\).
Intersecting with the domain: \((0, 81)\).
Exercise 11 — level ★★★☆☆
Solve:
\[(\log_2 x)^2 - 5\log_2 x + 6 \leq 0\]
Answer
\[[4,8]\]
Solution
Domain. \(x > 0\).
Let \(t = \log_2 x\): \(t^2 - 5t + 6 = (t-2)(t-3) \leq 0\).
The expression is non-positive between the roots: \(2 \leq t \leq 3\).
Therefore \(4 \leq x \leq 8\).
Intersecting with the domain: \([4, 8]\).
Exercise 12 — level ★★★☆☆
Solve:
\[(\log_3 x)^2 - 1 > 0\]
Answer
\[\left(0,\frac{1}{3}\right)\cup(3,+\infty)\]
Solution
Domain. \(x > 0\).
Let \(t = \log_3 x\): \(t^2 - 1 > 0 \iff (t-1)(t+1) > 0\).
Solutions: \(t < -1\) or \(t > 1\).
Therefore \(x < \frac{1}{3}\) or \(x > 3\).
Intersecting with the domain: \(\left(0,\frac{1}{3}\right)\cup(3,+\infty)\).
Exercise 13 — level ★★★☆☆
Solve:
\[2(\log_2 x)^2 - 3\log_2 x - 2 \geq 0\]
Answer
\[\left(0,\frac{1}{\sqrt{2}}\right]\cup[4,+\infty)\]
Solution
Domain. \(x > 0\).
Let \(t = \log_2 x\): \(2t^2 - 3t - 2 = (2t+1)(t-2) \geq 0\).
Solutions: \(t \leq -\frac{1}{2}\) or \(t \geq 2\).
Therefore \(x \leq 2^{-1/2} = \frac{1}{\sqrt{2}}\) or \(x \geq 4\).
Intersecting with the domain: \(\left(0,\frac{1}{\sqrt{2}}\right]\cup[4,+\infty)\).
Exercise 14 — level ★★★☆☆
Solve:
\[\log_2(x^2-5x+6)\geq 1\]
Answer
\[(-\infty,1]\cup[4,+\infty)\]
Solution
Domain. \(x^2-5x+6 > 0 \iff (x-2)(x-3)>0 \iff (-\infty,2)\cup(3,+\infty)\).
Base \(2>1\): \(x^2-5x+6 \geq 2 \iff x^2-5x+4 \geq 0 \iff (x-1)(x-4)\geq 0\).
Solutions: \(x\leq 1\) or \(x\geq 4\).
Intersecting with the domain: \((-\infty,1]\cup[4,+\infty)\).
Exercise 15 — level ★★★☆☆
Solve:
\[\log_3(x^2-4)<2\]
Answer
\[(-\sqrt{13},-2)\cup(2,\sqrt{13})\]
Solution
Domain. \(x^2-4 > 0 \iff (-\infty,-2)\cup(2,+\infty)\).
Base \(3>1\): \(x^2-4 < 9 \iff x^2 < 13 \iff -\sqrt{13} < x < \sqrt{13}\).
Intersecting with the domain: \((-\sqrt{13},-2)\cup(2,\sqrt{13})\).
Exercise 16 — level ★★★★☆
Solve:
\[\log_2(x+1)+\log_2(x-2)\leq \log_2(2x+3)\]
Answer
\[\left(2,\dfrac{3+\sqrt{29}}{2}\right]\]
Solution
Domain. \(x+1>0\), \(x-2>0\), \(2x+3>0\); the most restrictive condition is \(x>2\).
Logarithm product rule: \(\log_2[(x+1)(x-2)] \leq \log_2(2x+3)\).
Base \(2>1\): \((x+1)(x-2) \leq 2x+3 \iff x^2 - 3x - 5 \leq 0\).
Roots: \(\frac{3 \pm \sqrt{29}}{2}\). The solution lies between the roots.
Intersecting with \(x>2\): \(\left(2, \frac{3+\sqrt{29}}{2}\right]\).
Exercise 17 — level ★★★★☆
Solve:
\[\log_{1/2}(x^2-1)\geq -2\]
Answer
\[[-\sqrt{5},-1)\cup(1,\sqrt{5}]\]
Solution
Domain. \(x^2-1 > 0 \iff x \in (-\infty,-1)\cup(1,+\infty)\).
Base \(\frac{1}{2} < 1\), decreasing function, direction reversed:
\[x^2-1 \leq \left(\frac{1}{2}\right)^{-2} = 4 \iff x^2 \leq 5 \iff -\sqrt{5} \leq x \leq \sqrt{5}.\]
Intersecting with the domain: the values \(\pm 1\) are excluded since the argument vanishes there, while \(x = -\sqrt{5}\) and \(x = \sqrt{5}\) belong to the domain (the argument equals 4) and satisfy the inequality with equality. The solution set is therefore \([-\sqrt{5},-1)\cup(1,\sqrt{5}]\).
Exercise 18 — level ★★★★☆
Solve:
\[\log_2(x-1)<\log_4(3x+1)\]
Answer
\[(1,5)\]
Solution
Domain. \(x>1\).
\(\log_4(3x+1) = \frac{1}{2}\log_2(3x+1)\).
This leads to: \((x-1)^2 < 3x+1 \iff x^2 - 5x < 0 \iff x(x-5) < 0 \iff 0 < x < 5\).
Intersecting with the domain: \((1,5)\).
Exercise 19 — level ★★★★☆
Solve:
\[\log_3(x+1)+\log_3(5-x)\geq 1\]
Answer
\[[2-\sqrt{6},\ 2+\sqrt{6}]\]
Solution
Domain. \(-1 < x < 5\).
Logarithm product rule: \(\log_3[(x+1)(5-x)] \geq 1 \iff (x+1)(5-x) \geq 3\).
\(-x^2 + 4x + 2 \geq 0 \iff x^2 - 4x - 2 \leq 0\).
Roots: \(2 \pm \sqrt{6}\). The solution lies between the roots and is entirely contained within the domain.
Therefore \([2-\sqrt{6}, 2+\sqrt{6}]\).
Exercise 20 — level ★★★★★
Solve:
\[\log_2(x^2-4x+3)\leq \log_2(2x+3)\]
Answer
\[[0,1)\cup(3,6]\]
Solution
Domain. \(x^2-4x+3 > 0\) and \(2x+3 > 0\) \(\implies\) \(\left(-\frac{3}{2},1\right)\cup(3,+\infty)\).
Base \(2>1\): \(x^2-4x+3 \leq 2x+3 \iff x^2-6x \leq 0 \iff x(x-6)\leq 0\).
Solution: \(0 \leq x \leq 6\).
Intersecting with the domain: \([0,1)\cup(3,6]\).