Question

Find the real value of X

$\sf{(x²-7x+11)^{x²-13x+42} = 1}$


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Answer 1

Step-by-step explanation:

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$\sf{(x²-7x+11)^{x²-13x+42} = 1}$

==> (x² − 13x + 42)⋅ln( x² − 7x + 11) = 0

Because ln(z)=0 has no solution we stuck with

==> x²−13x+42=0

we use the small quadratic formula :-

==> x²+px+q=0 ⟹ x = −p/2 ± √p²/4 - q

and find:

x=6∨7

On the other hand we can also use :

∀r ∈ R(1r=1)

So we when we look at :-

==> x²−7x+11 = 1

we find

==> x=5 and x=2

And one more point :-

(−1) k = 1 if k is an even integer, so checking:

==> x²−7x+11=−1

concludes

==> x=3∨4

Now plugin them into:

==>> x²−13x+42 gets us 12∨6 both are even integers so both of them work.

So we have:

2, 3, 4, 5, 6 and 7 as solutions.

Answer 2

Step-by-step explanation:

Step-by-step explanation:

$\huge{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

$\sf{(x²-7x+11)^{x²-13x+42} = 1}$

==> (x² − 13x + 42)⋅ln( x² − 7x + 11) = 0

Because ln(z)=0 has no solution we stuck with

==> x²−13x+42=0

we use the small quadratic formula :-

==> x²+px+q=0 ⟹ x = −p/2 ± √p²/4 - q

and find:

x=6∨7

On the other hand we can also use :

∀r ∈ R(1r=1)

So we when we look at :-

==> x²−7x+11 = 1

we find

==> x=5 and x=2

And one more point :-

(−1) k = 1 if k is an even integer, so checking:

==> x²−7x+11=−1

concludes

==> x=3∨4

Now plugin them into:

==>> x²−13x+42 gets us 12∨6 both are even integers so both of them work.

So we have:

2, 3, 4, 5, 6 and 7 as solutions.