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 determine the count of unique positive integer solutions for the equation \((x^2-7x+11)^{(x^2-13x+42)}=1\), analyze the conditions under which an expression of the form \(a^b=1\) is satisfied:

• The equality \(a^b=1\) is true under the following circumstances:
  1. When the base \(a\) equals 1 (regardless of the exponent \(b\)).
  2. When the exponent \(b\) equals 0 (provided the base \(a\) is not 0).
  3. When the base \(a\) equals -1, and the exponent \(b\) is an even integer.

Condition 1: Set the base to 1: \(x^2-7x+11=1\)

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

Factoring this quadratic equation yields \((x-5)(x-2)=0\), resulting in solutions \(x=5\) and \(x=2\).

Condition 2: Set the exponent to 0: \(x^2-13x+42=0\)

Factoring this equation gives \((x-6)(x-7)=0\), yielding solutions \(x=6\) and \(x=7\). Note that for these values, the base \(x^2-7x+11\) is non-zero.

Condition 3: Set the base to -1 and check for an even exponent: \(x^2-7x+11=-1\)

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

Factoring this equation results in \((x-3)(x-4)=0\), giving potential solutions \(x=3\) and \(x=4\).

Verification for Condition 3: Confirm that the exponent \(x^2-13x+42\) is even for \(x=3\) and \(x=4\):

• For \(x=3\), \(x^2-13x+42 = 9 - 39 + 42 = 12\), which is even.
• For \(x=4\), \(x^2-13x+42 = 16 - 52 + 42 = 6\), which is also even.

The distinct positive integer solutions identified are \(x=2, 3, 4, 5, 6, 7\).

Therefore, there are 6 distinct positive integer solutions.