If a, b, c are positive real numbers such that a > 8 ≥ b ≥ c ≥ 2, and log₂(a + b) + log₂c = 6, then the greatest possible integer value of a is:
Answer and solution
Correct answer: B
Solution
Answer: B — 30
1. Combine the two logarithms:
2. log₂(a + b) + log₂c = log₂[c(a + b)]
3. Therefore,
4. log₂[c(a + b)] = 6
5. So,
6. c(a + b) = 2⁶ = 64
7. Rearranging:
8. a = 64/c − b
9. We need the greatest value of a. Since c and b are both at least 2, the smallest possible values are c = 2 and b = 2.
10. Putting these values in the expression:
11. a = 64/2 − 2
a = 32 − 2
a = 30
12. This also satisfies the condition a > 8.
13. Answer: 30.
Why this method works:
The upper bound follows without calculus: c≥2 gives 64/c≤32, and b≥2 subtracts at least 2. Therefore a≤30. A bound alone is insufficient unless a legal example reaches it.
Independent check and common trap:
The equality choice a=30, b=2, c=2 respects a>8≥b≥c≥2, and log₂32+log₂2=5+1=6. Thus 30 is attainable and is the greatest integer, not merely an upper estimate.
Apply: c(a+b)=64. Since b, c≥2, a=64/c−b≤32−2=30. Equality at b=c=2 is legal.
Check: The equality choice a=30, b=2, c=2 respects a>8≥b≥c≥2, and log₂32+log₂2=5+1=6. Thus 30 is attainable and is the greatest integer, not merely an upper estimate.
1. Combine the two logarithms:
2. log₂(a + b) + log₂c = log₂[c(a + b)]
3. Therefore,
4. log₂[c(a + b)] = 6
5. So,
6. c(a + b) = 2⁶ = 64
7. Rearranging:
8. a = 64/c − b
9. We need the greatest value of a. Since c and b are both at least 2, the smallest possible values are c = 2 and b = 2.
10. Putting these values in the expression:
11. a = 64/2 − 2
a = 32 − 2
a = 30
12. This also satisfies the condition a > 8.
13. Answer: 30.
Why this method works:
The upper bound follows without calculus: c≥2 gives 64/c≤32, and b≥2 subtracts at least 2. Therefore a≤30. A bound alone is insufficient unless a legal example reaches it.
Independent check and common trap:
The equality choice a=30, b=2, c=2 respects a>8≥b≥c≥2, and log₂32+log₂2=5+1=6. Thus 30 is attainable and is the greatest integer, not merely an upper estimate.
G Strategy shortcut
Spot: Algebra – Logarithms.Apply: c(a+b)=64. Since b, c≥2, a=64/c−b≤32−2=30. Equality at b=c=2 is legal.
Check: The equality choice a=30, b=2, c=2 respects a>8≥b≥c≥2, and log₂32+log₂2=5+1=6. Thus 30 is attainable and is the greatest integer, not merely an upper estimate.
G Strategy: Spot: Algebra – Logarithms. Apply: c(a+b)=64. Since b, c≥2, a=64/c−b≤32−2=30. Equality at b=c=2 is legal. Check: The equality choice a=30, b=2, c=2 respects a>8≥b≥c≥2, and log₂32+log₂2=5+1=6. Thus 30 is attainable and is the greatest integer, not merely an upper estimate.