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(1/81)^(x) + (1/9)^(x) = 1[(1/9)²]^(x) + (1/9)^(x) = 1[(1/9)^(x)]² + (1/9)^(x) = 1 → let: a = (1/9)^(x)a² + a = 1a² + a - 1 = 0Δ = (1)² - (4 * - 1) = 5a = (- 1 ± √5)/2 → recall: a = (1/9)^(x)(1/9)^(x) = (- 1 ± √5)/2 → we keep on,ly the positive value(1/9)^(x) = (- 1 + √5)/2Ln[(1/9)^(x)] = Ln[(- 1 + √5)/2]x.Ln(1/9) = Ln[(- 1 + √5)/2]x = Ln[(- 1 + √5)/2] / Ln(9) → to go furtherx = [Ln(- 1 + √5) - Ln(2)] / 2.Ln(3)
/* 81^x , 81^x-9^x-1=0 , let u=9^x , u^2-u-1=0 , u=(1+V5)/2 , / (1-V5)/2 , not a solu , < 0 , / , 9^x=(1+V5)/2 , x=log((1+V5)/2)/log9 , test , (1/81)^x+(1/9)^x=0.381966+0.618034 , --> 1 , OK ,
(1/81)^(x) + (1/9)^(x) = 1
[(1/9)²]^(x) + (1/9)^(x) = 1
[(1/9)^(x)]² + (1/9)^(x) = 1 → let: a = (1/9)^(x)
a² + a = 1
a² + a - 1 = 0
Δ = (1)² - (4 * - 1) = 5
a = (- 1 ± √5)/2 → recall: a = (1/9)^(x)
(1/9)^(x) = (- 1 ± √5)/2 → we keep on,ly the positive value
(1/9)^(x) = (- 1 + √5)/2
Ln[(1/9)^(x)] = Ln[(- 1 + √5)/2]
x.Ln(1/9) = Ln[(- 1 + √5)/2]
x = Ln[(- 1 + √5)/2] / Ln(9) → to go further
x = [Ln(- 1 + √5) - Ln(2)] / 2.Ln(3)
/* 81^x , 81^x-9^x-1=0 , let u=9^x , u^2-u-1=0 , u=(1+V5)/2 , / (1-V5)/2 , not a solu , < 0 , / , 9^x=(1+V5)/2 ,
x=log((1+V5)/2)/log9 , test , (1/81)^x+(1/9)^x=0.381966+0.618034 , --> 1 , OK ,