Question

Find the real value of X

$\sf{(x²-7x+11)^{x²-13x+42} = 1}$

wrong Answer= I'd report​

Answer 1

Answer:

For the whole expression to be equal to 1, either the expression in the power is zero or the expression in the base must be equal to 1

(Anything)^0=1

1^(anything) = 1

Let’s solve separately

x^2–7x+11=1 =>x^2–7x+10=0=>(x-2)(x-5)=0

x1=2, x2=5

now let’s solve the power equation

x^2–13x+42=0 =>(x-7)(x-6)=0

x3=6, x4=7

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

this question's answer is x = 7 , 6 , 5 , 2