Since \(2^x\) is increasing, \(2^x\le3\) is equivalent to \(x\le\log_2 3\). Answer: A.
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2
Task
Numeric answer
Solve \(\lg\dfrac4{2x-3}\ge0\). Enter the greatest solution. If no greatest solution exists, enter \(100\).
Solution
Since \(\lg t\ge0\) when \(t\ge1\), we need \(\dfrac4{2x-3}\ge1\), with the logarithm domain condition \(2x-3\gt0\). Thus \(0\lt2x-3\le4\), so \(1.5\lt x\le3.5\). The greatest solution is \(3.5\).
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3
Task
One correct answer
Solve \(\log_3x\lt-1\).
Solution
Write \(-1=\log_3\dfrac13\). Since the base is greater than \(1\), and \(x\gt0\), we get \(0\lt x\lt\dfrac13\). Answer: D.
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4
Task
One correct answer
Solve \(\log_{\frac15}x\gt2\).
Solution
Write \(2=\log_{1/5}\dfrac1{25}\). Since the base is less than \(1\), the inequality reverses, giving \(x\lt\dfrac1{25}\). With \(x\gt0\), \(x\in(0;\dfrac1{25})\). Answer: C.
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5
Task
Numeric answer
Solve \(x^2+2^{\log_2(-2x)}-15\lt0\). Enter the sum of all integer solutions.
Solution
Domain: \(x\lt0\). Then \(2^{\log_2(-2x)}=-2x\), so \((x+3)(x-5)\lt0\), giving \(-3\lt x\lt5\). Intersecting with the domain yields \(-3\lt x\lt0\). The integer solutions are \(-2,-1\), whose sum is \(-3\).
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6
Task
Numeric answer
Find the number of all integer solutions of \(\log_{\frac14}(x^2+6x)\ge-2\). If there are infinitely many, enter \(100\).
Solution
Domain: \(x^2+6x\gt0\). Since the base is less than \(1\), we also need \(x^2+6x\le16\). Thus \(0\lt x^2+6x\le16\), giving \(x\in[-8,-6)\cup(0,2]\). The integer solutions are \(-8,-7,1,2\), so there are \(4\).
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7
Task
One correct answer
Solve \(\log_{0.5}(x-1)\gt2\).
Solution
Domain: \(x\gt1\). Since the base \(0.5\lt1\), \(\log_{0.5}(x-1)\gt2\) gives \(x-1\lt0.25\). Thus \(1\lt x\lt1.25\). Answer: A.
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8
Task
One correct answer
Solve \(3+\log_2x\ge0\).
Solution
The inequality is equivalent to \(\log_2x\ge-3=\log_2\dfrac18\). Since the base is greater than \(1\), \(x\ge\dfrac18\). Answer: A.
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9
Task
One correct answer
Solve \(\log_{0.1}10\lt\log_{0.1}x\).
Solution
Domain: \(x\gt0\). Since the base is less than \(1\), the order reverses, giving \(10\gt x\). Hence \(0\lt x\lt10\). Answer: B.
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10
Task
One correct answer
Solve \(\log_{0.5}5\lt\log_{0.5}x\).
Solution
Domain: \(x\gt0\). Since the base \(0.5\lt1\), we get \(5\gt x\). Hence \(x\in(0,5)\). Answer: B.
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11
Task
One correct answer
Solve \(\log_{0.7}x\gt1\).
Solution
Domain: \(x\gt0\). Since the base \(0.7\lt1\), the logarithmic function is decreasing. Thus \(x\lt0.7\), so \(x\in(0,0.7)\). Answer: E.
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12
Task
One correct answer
Solve \(\log_{0.9}(3x)\gt2\).
Solution
Domain: \(x\gt0\). Since the base is less than \(1\), \(\log_{0.9}(3x)\gt2\) gives \(3x\lt0.81\), hence \(x\lt0.27\). Therefore \(x\in(0,0.27)\). Answer: E.
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13
Task
One correct answer
Solve \(\log_{0.3}(x+3)\gt\log_{0.3}4\).
Solution
Domain: \(x\gt-3\). Since \(0.3\lt1\), the logarithmic function is decreasing, so \(x+3\lt4\), hence \(x\lt1\). Thus \(x\in(-3,1)\). Answer: E.
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14
Task
One correct answer
Solve \(\log_{0.2}(x+4)\gt\log_{0.2}5\).
Solution
Domain: \(x\gt-4\). Since the base is less than \(1\), we get \(x+4\lt5\), so \(x\lt1\). Thus \(x\in(-4,1)\). Answer: C.
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15
Task
One correct answer
Solve \(\log_3(1-2x)\lt2\).
Solution
Domain: \(x\lt\dfrac12\). Since \(3\gt1\), we get \(1-2x\lt9\), hence \(x\gt-4\). Thus \(x\in(-4,\dfrac12)\). Answer: B.
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16
Task
One correct answer
Using the graph, solve \(\log_2x\lt b\).
Solution
The function \(y=\log_2x\) is increasing. The line \(y=b\) intersects it at \(x=2^b\). The graph lies below this line for \(0\lt x\lt2^b\). Answer: A.
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17
Task
One correct answer
Solve \(\log_3(2x-1)\lt2\).
Solution
Domain: \(x\gt0.5\). Since \(3\gt1\), we get \(2x-1\lt9\), so \(x\lt5\). Thus \(x\in(0.5,5)\). Answer: A.
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18
Task
One correct answer
Solve \(\log_4(2x-2)\lt2\).
Solution
Domain: \(x\gt1\). Since \(4\gt1\), we get \(2x-2\lt16\), hence \(x\lt9\). Thus \(x\in(1,9)\). Answer: B.
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19
Task
One correct answer
Solve \(3+\log_2x\lt\log_2 48\).
Solution
Domain: \(x\gt0\). Since \(3=\log_2 8\), we get \(\log_2(8x)\lt\log_2 48\). Thus \(8x\lt48\), so \(x\lt6\). Hence \(x\in(0,6)\). Answer: D.
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20
Task
Matching
Solve inequalities 1–4 and match each with its solution set A–E.
А
\((-\infty;2)\) Б
\((-2;2)\) В
\((0;2)\) Г
\((-\infty;-2)\) Д
\((2;+\infty)\)
Solution
1) \(x\gt2\) — E. 2) \(x\lt-2\) — D. 3) \(0\lt x\lt2\) — C. 4) \(-2\lt x\lt2\) — B. Thus: **1–E, 2–D, 3–C, 4–B**.
