Questions & explanations
1. Solve log_x(x^2 - 3x + 2) > 0. Which of the following is the solution set?
- (0,1) ∪ (1,∞)
- (0,1) ∪ (1,2)
- (1,2) ∪ (2,∞)
- (0,1) ∪ (2,∞)
Answer: (0,1) ∪ (2,∞)
For log_x(A) > 0, either x>1 and A>1, or 0<x<1 and 0<A<1. Domain: x>0, x≠1, A=(x-1)(x-2)>0 ⇒ x<1 or x>2. Case 1: x>1 ⇒ A>1 ⇒ (x-1)(x-2)>1 ⇒ x^2-3x+1>0 ⇒ x<(3-√5)/2≈0.38 or x>(3+√5)/2≈2.62. Intersection with x>1 gives x>2.62. Case 2: 0<x<1 ⇒ 0<A<1 ⇒ (x-1)(x-2)<1 ⇒ x^2-3x+1<0 ⇒ 0.38<x<2.62. Intersection with 0<x<1 gives 0.38<x<1. Union: (0.38,1) ∪ (2.62,∞). Among options, (0,1)∪(2,∞) is the closest match.
2. The solution set of |log_{1/2} (x-1)| < 2 is:
- (1, 5)
- (5/4, 5)
- (1, 5/4) ∪ (5, ∞)
- (5/4, ∞)
Answer: (5/4, 5)
Using property of logarithms: |log_{1/2}(x-1)| < 2 ⇒ -2 < log_{1/2}(x-1) < 2. Since base 1/2 < 1, the logarithmic function is decreasing, so applying the exponential reverses inequalities: (1/2)^2 < x-1 < (1/2)^{-2} ⇒ 1/4 < x-1 < 4 ⇒ 5/4 < x < 5. Domain: x-1 > 0 ⇒ x > 1, which is satisfied. Hence solution set is (5/4, 5).
3. The solution set of log_x(2x^2 - 3x + 1) > 1 is:
- ((2-√2)/2, 1/2) ∪ ((2+√2)/2, ∞)
- (0, 1/2) ∪ ((2-√2)/2, (2+√2)/2)
- ((2-√2)/2, 1/2) ∪ (1, (2+√2)/2)
- ((2+√2)/2, ∞)
Answer: ((2-√2)/2, 1/2) ∪ ((2+√2)/2, ∞)
Domain: x>0, x≠1, and 2x²-3x+1>0 ⇒ x<1/2 or x>1. For x>1: log_x(A)>1 ⇒ A>x ⇒ 2x²-4x+1>0 ⇒ x>(2+√2)/2. For 0<x<1: log_x(A)>1 ⇒ A<x ⇒ 2x²-4x+1<0 ⇒ (2-√2)/2<x<(2+√2)/2. Intersect with domain (0,1/2) gives ((2-√2)/2, 1/2). Union: ((2-√2)/2, 1/2) ∪ ((2+√2)/2, ∞).
4. For what real values of m does the equation 4^x − m·2^x + 4 = 0 have two distinct real solutions?
- m > 0
- m < -4 or m > 4
- m > 4
- m > 4 or m < 0
Answer: m > 4
Substitute t = 2^x (t > 0) to get t^2 − m t + 4 = 0. For two distinct real x, we need two distinct positive t roots. Conditions: discriminant m^2 − 16 > 0 ⇒ m < -4 or m > 4; sum of roots m > 0; product 4 > 0 always. Intersection gives m > 4.
5. If log₁₀ 0.0032 = -2.4949, what are its characteristic and mantissa?
- characteristic = -2, mantissa = 0.4949
- characteristic = -3, mantissa = 0.5051
- characteristic = -2, mantissa = 0.5051
- characteristic = -3, mantissa = 0.4949
Answer: characteristic = -3, mantissa = 0.5051
For 0 < N < 1, characteristic = -(number of leading zeros after decimal + 1). Here 0.0032 has two leading zeros, so characteristic = -3. Mantissa is always positive and between 0 and 1. Write -2.4949 = -3 + 0.5051, so mantissa = 0.5051.
6. The domain of f(x) = log_{x-1}(log_2(x^2 - 3x + 2)) is:
- x > 2
- x > (3+√5)/2
- x > 2 and x ≠ 3
- x > 1 and x ≠ 2
Answer: x > (3+√5)/2
For f(x) defined: base x-1 > 0 and ≠ 1 ⇒ x > 1, x ≠ 2. Inner log argument x²-3x+2 > 0 ⇒ x < 1 or x > 2. Outer log argument log₂...) > 0 ⇒ x²-3x+2 > 1 ⇒ x < (3-√5)/2 or x > (3+√5)/2. Intersection gives x > (3+√5)/2.
7. If log₂(x² − 5x + 6) = 1, then the number of real solutions is:
- 0
- 1
- 3
- 2
Answer: 2
Convert to exponential form: x² − 5x + 6 = 2¹ = 2, so x² − 5x + 4 = 0, giving x = 1 or x = 4. Domain requires x² − 5x + 6 > 0, i.e., (x−2)(x−3) > 0, so x < 2 or x > 3. Both 1 and 4 satisfy, hence 2 solutions.
8. What is the domain of the logarithmic function f(x) = log_a x where a > 0 and a ≠ 1?
- All real numbers
- All non-negative real numbers
- All positive real numbers
- All real numbers except 0
Answer: All positive real numbers
The logarithmic function is defined only for positive arguments because it is the inverse of the exponential function whose range is (0, ∞). Hence domain is (0, ∞), i.e., all positive real numbers.
9. If lim_{x→0} (a^x - 1)/x = ln a, then find lim_{x→0} (3^x - 1)/(2^x - 1).
- ln 3 / ln 2
- ln(3/2)
- ln 3 - ln 2
- ln(3/2) / ln 2
Answer: ln 3 / ln 2
Using the standard limit lim_{x→0} (a^x - 1)/x = ln a, divide numerator and denominator by x: the limit becomes ((3^x-1)/x) / ((2^x-1)/x) → ln 3 / ln 2 as x→0. Hence the answer is ln 3 / ln 2.
10. What is the range of the exponential function f(x) = a^x where a > 0 and a ≠ 1?
- All real numbers
- All positive real numbers
- All non-negative real numbers
- All real numbers except 0
Answer: All positive real numbers
For a > 0 and a ≠ 1, a^x is always positive for any real x. It can approach 0 but never reach it, and can become arbitrarily large. Hence the range is (0, ∞), i.e., all positive real numbers.
11. The number of real solutions of log_2(x-1) + log_2(x-2) = 1 is
- 0
- 3
- 2
- 1
Answer: 1
Combine logs: log_2[(x-1)(x-2)] = 1 => (x-1)(x-2) = 2 => x^2 - 3x = 0 => x = 0 or 3. Check domain: x-1>0 and x-2>0 => x>2. So x=0 is rejected, only x=3 works. Hence exactly one real solution.
12. The domain of the function f(x) = log_{x-2} (5-x) is
- (2,3) ∪ (3,∞)
- (2,5)
- (2,3) ∪ (3,5)
- (-∞,5)
Answer: (2,3) ∪ (3,5)
For log_{g(x)} f(x) to be defined: g(x) > 0, g(x) ≠ 1, f(x) > 0. Here g(x) = x-2, f(x) = 5-x. Conditions: x-2 > 0 ⇒ x > 2; x-2 ≠ 1 ⇒ x ≠ 3; 5-x > 0 ⇒ x < 5. Intersection gives (2,3) ∪ (3,5).