中.3 數學題一條

(x+4)3/5=(5x-1)2/3
((x+4)3/5)15=((5x-1)2/3)15
(x+4)9=(5x-1)10
(5x-1)10 - (x+4)9 = 0
Expanding the equation, excluding the complex roots,
x = - 0.43 or x = 1.1 (corr. to 2 d.p.)

(x+4)^3/5=(5x-1)^2/3
(3/5)log(x+4)=(2/3)log(5x-1)
(9/10)log(x+4)=log(5x-1)
(9/10)log(x+4)-log(5x-1)=0
let P(x)=(9/10)log(x+4)-log(5x-1)

so, x must be greater than 1/5
using bisection method
first take x =1 , P(x)>0
and take x =1.1 , P(x)<0

then take x =1.05 , P(x)>0
so x fell between 1.05 and 1.1

then take x =1.075 , P(x)<0
so x fell between 1.05 and 1.075

then take x =1.0625 , P(x)<0
so x fell between 1.05 and 1.0625

then take x =1.05625 , P(x)>0
so x fell between 1.05625 and 1.0625

then take x =1.059375 , P(x)>0
so x fell between 1.059375 and 1.0625

x=1.06 (corrected to 2 d.p.)