Question

How many distinct positive integer-valued solutions exist to the equation \((x^2-7x+11)^{(x^2-13x+42)}=1\)?

• A. 6
• B. 8
• C. 2
• D. 4

Answer 1

The Correct Option is A

To find the number of distinct positive integer solutions to the equation \((x^2-7x+11)^{(x^2-13x+42)}=1\), consider the properties of exponents:

• The expression \(a^b=1\) holds true if:
  1. \(a=1\) (irrespective of \(b\))
  2. \(b=0\) (for any \(a \neq 0\))
  3. \(a=-1\) and \(b\) is even 

Step 1: Solve \(x^2-7x+11=1\)

\(x^2-7x+10=0\)

Factoring, \((x-5)(x-2)=0\), so \(x=5\text{ or }x=2\).

Step 2: Solve \(x^2-13x+42=0\)

Factor as \((x-6)(x-7)=0\), giving solutions \(x=6\text{ or }x=7\).

Step 3: Solve \(x^2-7x+11=-1\) when \(x^2-13x+42\) is even:

\(x^2-7x+12=0\)

Factor as \((x-3)(x-4)=0\), thus \(x=3\text{ or }x=4\).

Verification: Check the parity of \(x^2-13x+42\):

• For \(x=3\) and \(x=4\), the calculation yields even results for \(x^2-13x+42\).

Total Distinct Solutions: \(x=2, 3, 4, 5, 6, 7\)

Hence, there are 6 distinct positive integer solutions.

Answer 2

We are given: \[ (x^2 - 7x + 11)^{(x^2 - 13x + 42)} = 1 \] Now, for any expression of the form \( a^b = 1 \), it can be true in the following cases:

1. \( a = 1 \) and \( b \in \mathbb{R} \) 
2. \( a = -1 \) and \( b \) is even
3. \( b = 0 \) and \( a \ne 0 \)

Let us analyze each of these cases: 
Case 1: Exponent is 0 \[ x^2 - 13x + 42 = 0 \] \[ (x - 6)(x - 7) = 0 \Rightarrow x = 6, 7 \] Now check base for these values: - At \( x = 6 \), base = \( x^2 - 7x + 11 = 36 - 42 + 11 = 5 \ne 0 \) - At \( x = 7 \), base = \( 49 - 49 + 11 = 11 \ne 0 \) So both values are valid. 
Case 2: Base is 1 \[ x^2 - 7x + 11 = 1 \Rightarrow x^2 - 7x + 10 = 0 \] \[ (x - 2)(x - 5) = 0 \Rightarrow x = 2, 5 \] 
Case 3: Base is -1 and exponent is even \[ x^2 - 7x + 11 = -1 \Rightarrow x^2 - 7x + 12 = 0 \] \[ (x - 3)(x - 4) = 0 \Rightarrow x = 3, 4 \] Now check if the exponent is even at these values: - At \( x = 3 \), exponent = \( 9 - 39 + 42 = 12 \) → even ✅ - At \( x = 4 \), exponent = \( 16 - 52 + 42 = 6 \) → even ✅ Hence, all values are valid: \[ x = 2, 3, 4, 5, 6, 7 \] Total number of valid real values of \( x \) is: \[ \boxed{6} \] So, the correct answer is option (A): 6