Question

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

Answer 1

(x² - 7x + 11)^(x² - 13x + 42) = 1

$({x}^{2} - 7x + 11 ){}^{ {x}^{2} - 13x + 42} = ( {x}^{2} - 7x + 11) {}^{0}$

Since, anything raised to the power 0 = 1

Now since the bases are same, we can compare the powers.

=>
${x}^{2} - 13x + 42 = 0$

Now, factorise the LHS by splitting the middle term,

x² - 13x + 42 = 0

=> x² - 7x - 6x + 42 = 0

=> x(x - 7) - 6(x - 7) = 0

=> (x - 6)(x - 7) = 0

=> (x - 6) = 0 ÷ (x - 7)

=> (x - 6) = 0

=> x = 6

Similarly,

(x - 7)(x - 6) = 0

=> (x - 7) = 0 ÷ (x - 6)

=> (x - 7) = 0

=> x = 7

Hence, your answer is, x = 6 or x = 7

Answer 2

Here is  One  more  Possible  Solution

Answer:

x = 2 or x = 5

Step-by-step explanation:

We  have,

$(x^2-7x+11)^{(x^2-13x+42)}=1\\\;\\\text{Taking log of both sides}\\\;\\(x^2-13x+42)\log(x^2-13x+11)=\log1\\\;\\(x^2-13x+42)\log(x^2-13x+11)=0\\\;\\\log(x^2-13x+11)=\frac{0}{(x^2-13x+42)}\\\;\\\log(x^2-13x+11)=0\\\;\\\log(x^2-13x+11)=\log1\\\;\\x^2-13x+11=1\\\;\\x^2-13x+10=0\\\;\\x^2-5x-2x+10=0\\\;\\x(x-5)-2(x-5)=0\\\;\\(x-2)(x-5)=0\\\;\\Hence,\\\;\\\textbf{x=2\;\;or\;\;x=5}\\\\Note:-\\\;\\log1=0\\\;\\\log{a^m}=m\log{a}$