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.
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(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
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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
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Since ∀a∈ℝ*, 0ª=0≠1, assuming x²–8x+15=0 is not a credible hypothesis.
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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.
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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)}
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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.
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Congrats to
► Ray for seeing the tricky possibility (-1)²ⁿ=1
► D.W. and Minh for seeing the trap 0º
► Johan for the try.
:-)