[Risolto] Equazione esponenziale n. 47

Ciao @beppe

8^(x - 2)·12^(1/3) = 9^((4 + x)/6)/(32·81^((1 - 2·x)/3))

2^(3·(x - 2))·12^(1/3) = 3^(2·(4 + x)/6)/(32·3^(4·(1 - 2·x)/3))

2^(3·x - 6)·12^(1/3) = 3^(x/3 + 4/3)/(32·3^(4/3 - 8·x/3))

2^(3·x)/2^6·12^(1/3) = 3^(x/3 + 4/3 - (4/3 - 8/3·x))/32

2^(3·x)/2^6·12^(1/3) = 3^(3·x)/32

2^(3·x)/3^(3·x) = 2^6/(32·12^(1/3))

(2/3)^(3·x) = 2/12^(1/3)

metto sotto segno di logaritmo i due membri:

3·x·LN(2/3) = LN(2/12^(1/3))

3·x = (LN(2) - 1/3·LN(12))/(LN(2) - LN(3))

3·x = (3·LN(2) - LN(12))/3/(LN(2) - LN(3))

3·x = (3·LN(2) - 2·LN(2) - LN(3))/3/(LN(2) - LN(3))

3·x = 1/3·((LN(2) - LN(3))/(LN(2) - LN(3)))

3·x = 1/3------> x = 1/9

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L'equazione
47) (8^(x - 2))*12^(1/3) = ((9^(4 + x))^(1/6))/(32*(81^(1 - 2*x))^(1/3))
si compone di tre subespressioni "a futticumpagnu" che anzitutto cerco di ridurre a forme più maneggevoli.
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A) (8^(x - 2))*12^(1/3) = (2^(3*x))*(2^(- 6))*(2^(2/3))*3^(1/3) =
= (2^(3*x - 16/3))*3^(1/3)
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B) (9^(4 + x))^(1/6) = 3^(2*(4 + x)/6) = 3^((x + 4)/3)
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C) 32*(81^(1 - 2*x))^(1/3) = (2^5)*3^(4*(1 - 2*x)/3)
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D) B/C = (3^((x + 4)/3))/((2^5)*3^(4*(1 - 2*x)/3)) =
= (1/2^5)*3^((x + 4)/3 - 4*(1 - 2*x)/3) =
= (1/2^5)*3^(3*x)
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47) (8^(x - 2))*12^(1/3) = ((9^(4 + x))^(1/6))/(32*(81^(1 - 2*x))^(1/3)) ≡
≡ (2^(3*x - 16/3))*3^(1/3) = (1/2^5)*3^(3*x) ≡
≡ (2^(3*x - 16/3))*2^5 = 3^(3*x)/3^(1/3) ≡
≡ 2^(3*x - 1/3) = 3^(3*x - 1/3) ≡
≡ (2/3)^(3*x - 1/3) = 1 ≡
≡ x = 1/9
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http://www.wolframalpha.com/input?i=solve%288%5E%28x-2%29%29*12%5E%281%2F3%29%3D%28%289%5E%284--x%29%29%5E%281%2F6%29%29%2F%2832*%2881%5E%281-2*x%29%29%5E%281%2F3%29%29for+x+real