Question

(x^2-7x+11)^(x^2-13x+42) =1 solve for x

there are 6 solutions...

Answer 1

Answer:

Thus the solutions are 2,3,4,5,6 and 7.

Step-by-step explanation:

Hi,

Consider aⁿ,

If a≠±1 and if a≠0, then aⁿ is always 1 whenever n is 0.----(1)

if a = 1 then aⁿ is always 1....(2)

if a = -1 then n should be even for aⁿ to be 1----(3)

Consider (x²-7x+11)^(x²-13x+42) =1

Case 1

If x²-13x+42 = 0

⇒ x² - 6x -7x + 42 = 0

⇒ (x - 6)(x-7) = 0

⇒ x = 6 or x = 7

If x = 6, x² - 7x + 11 = 6² -7.6 + 11 ≠ 0

Hence 6 is one solution

If x = 7, x² - 7x + 11 = 7² -7.7 + 11 ≠ 0

Hence 7 is one solution

Case 2:

If x² - 7x + 11 = 1

⇒x² -7x + 10 = 0

⇒(x - 2)(x - 5) = 0

⇒ x = 2 or x = 5 are another 2 solutions

If  x² - 7x + 11 = -1

⇒x² -7x + 12 = 0

⇒ (x - 3)(x - 4) = 0

=> x = 3 or x = 4

If x = 3, x² - 13x + 42 = 3² - 13*3 + 42 = 12 which is even

Hence , x = 3 is solution as stated in (3).

If x = 4, x² - 13x + 42 = 4² - 13*4 + 42 = 6 which is even

Hence , x = 4 is solution as stated in (3).

Thus the solutions are 2,3,4,5,6 and 7.

Hope, it helps !