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Tricky math problem if you are bored?
Hello,
DISCLAIMER:
I am not trying to get my homework done. This problem is provided as it is, and I do have the answer. So it is a challenge for anyone. Best answer will be awarded in approximatively two days. Have fun gals and guys!
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Find all the real solutions to:
(x² – 8x + 15)^(x³ – 9x) = 1
Regards,
Dragon.Jade :-)
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Bravo to all the competitors.
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There were two tricks to this problem.
► The first one was the hidden possibility that 0º could be reached by setting x=3. And that 0º is actually an undefined form.
► The second was that using aⁿ=1 ↔ n.ln(a)=0 would lead solvers to miss all solutions were a<0.
As the answers stand, nobody solved the problem perfectly.
= = = = = = = =
(x² – 8x + 15)^(x³ – 9x) = 1
ln[(x² – 8x + 15)^(x³ – 9x)] = 0
Assuming x²–8x+15>0, we can write:
(x³ – 9x).ln[(x² – 8x + 15) = 0
>
Then either
► x³ – 9x = x(x – 3)(x + 3) = 0 → x∈{-3; 0; 3}
But when x=3, x² – 8x + 15 = 9 – 24 + 15 = 0
Which contradicts the initial assumption, so x=3 is extraneous.
Or
► x² – 8x + 15 = 1
x² – 8x + 14 = (x – 4 – √2)(x – 4 + √2) = 0 → x = 4 ± √2
= = = = = = = =
Since ∀a∈ℝ*, 0ª=0≠1, assuming x²–8x+15=0 is not a credible hypothesis.
= = = = = = = =
Lastly, we need to consider the case where x²–8x+15<0.
The only way so that:
bª > 0
when b<0 is to have a to be an even integer.
>
So we have to consider the possibility
(-1)²ⁿ = 1
and solve
x² – 8x + 15 = -1
x² – 8x + 16 = 0
(x – 4)² = 0
x = 4
Then
x³ – 9x = 4³ – 9×4 = 64 – 36 = 28
which is indeed even.
So x=4 is also a solution.
Thus the complete solution set:
x∈{-3; 0; (4–√2); 4; (4+√2)}
= =
It is true that:
Lim(x→3) (x²–8x+15)^(x³–9x) = 1
However
x=3 is NOT a solution as 0º is undefined.
JUST LIKE
Lim(x→3) (x–3)/(x–3) = 1
But
x=3 is NOT a solution as any division by 0 is undefined.
= = = = = = = = = = = =
Congrats to
► Ray for seeing the tricky possibility (-1)²ⁿ=1
► D.W. and Minh for seeing the trap 0º
► Johan for the try.
:-)
4 réponses
- RayLv 6il y a 7 ansRéponse favorite
Very interesting.
(x^2 - 8x + 15)^(x^3 - 9·x) = 1
(x^3 - 9·x) * LN(x^2 - 8x + 15) = 0
So either (x^3 - 9x) = 0
x(x^2 - 9) = 0
x(x - 3)(x + 3) = 0 ... giving x = 0, x=3, x=-3 <<<<<<<<<<
or
LN(x^2 - 8x + 15) = 0
x^2 - 8x + 15 = 1
x^2 - 8x + 14 = 0
giving by quadratic formula x = 4 - √2 or x = 4 + √2 <<<<<<<<<<<
OR
(x^2 - 8x + 15) = -1 and x^3 - 9x is an even integer
So if (x^2 - 8x + 15) = -1
x^2 - 8x + 16 = 0
(x - 4)(x - 4) = 0 ... so x = 4
which gives (x^2 - 8x + 15)^(x^3 - 9·x)
= (4^2 - 32 + 15)^(4^3 - 36)
= -1 ^ 28
= 1 ... confirming that x = 4 is also a solution <<<<<<<<<<<
- JohanLv 5il y a 7 ans
(x² – 8x + 15)^(x³ – 9x) = 1
==> (x³ – 9x) = 0 , or (x² – 8x + 15) = 1
(x³ – 9x) = 0 :
x(x+3)(x-3) = 0
x = 0 , or x = -3, or x = 3
(x² – 8x + 15) = 1 :
x² – 8x + 16 = 2
(x - 4)² = 2
x-4 = √2 , or x-4 = -√2
x = 4 + √2, or x = 4 - √2
Checking each of these five found solutions:
x = 0 :
(x² – 8x + 15)^(x³ – 9x) =
= (0² – 8*0 + 15)^(0³ – 9*0)
= 15^0
= 1
checks out okay!
x = -3 :
(x² – 8x + 15)^(x³ – 9x) =
= ((-3)² – 8(-3) + 15)^((-3)³ – 9(-3))
= (9 + 24 + 15)^(-27 + 27)
= 48^0
= 1
checks out okay!
x = +3 :
(x² – 8x + 15)^(x³ – 9x) =
= (3² – 8*3 + 15)^(3³ – 9*3)
= (9 - 24 + 15)^(27 - 27)
= 0^0
= 1
checks out okay!
(see http://en.wikipedia.org/wiki/Exponentiation#Zero_t... )
x = 4 + √2 :
(x² – 8x + 15)^(x³ – 9x) =
= [ (4+√2)² – 8(4+√2) + 15 ]^((4+√2)³ – 9(4+√2))
= [ 18 + 8√2 - 32 - 8√2 + 15 ]^((4+√2)(18+8√2-9))
= [1]^((4+√2)(9+8√2))
= [1]^(36+9√2+32√2+16)
= [1]^(52 + 41√2)
= 1
checks out okay!
x = 4 - √2 :
(x² – 8x + 15)^(x³ – 9x) =
= [ (4-√2)² – 8(4-√2) + 15 ]^((4-√2)³ – 9(4-√2))
= [ 18 - 8√2 - 32 + 8√2 + 15 ]^((4-√2)(18-8√2-9))
= [1]^((4-√2)(9-8√2))
= [1]^(36-9√2-32√2+16)
= [1]^(52 - 41√2)
= 1
checks out okay!
So the solutions set is: {0, -3, +3, 4-√2, 4+√2}
- - - - - - - - - - - - - - - - - - - - - -
EDIT: Ouch! After seeing Ray's reply, I see that I missed one solution. Well done, Ray!
By the way, here are the (real-valued) solutions according to Wolfram Alpha:
- ?Lv 7il y a 7 ans
x³-9x = 0
x(x+3)(x-3) = 0
x = 0, 3, -3
3²-8·3+15 = 0, so x=0 is an extraneous solution.
x = 0, -3
- YoLv 4il y a 7 ans
interesting problem.
( (x-3) (x-5) )^( x (x - 3) (x + 3) ) = 1
certainly x = 0 and -3 will work,
let (x-3) (x-5) = 1
then x = 4 +/- sqrt(2)
all I have so for is 0, -3, and 4 +/- sqrt(2). Quite honestly, I'm not sure if there is more
