2題高一數

1.
2^2x-3-3‧2^x-2+1=0

(2^-3)(2^2x)-3(2^x)(2^-2)+1=0

(1/8)(2^2x)-3(2^x)(1/4)+1=0

(2^2x)-6(2^x)+8=0

(2^x-2)(2^x-4)=0

2^x=2 or 2^x=4

2^x=2^1 or 2^x=2^2

x=1 or x=2

2.
2logx^25-3log25^x=1

logx^50-log25^3x=1

log(x^50/5^6x)=1

x^50/5^6x=10

x^50-10(5^6x)=0

let f(x)=x^50-10(5^6x)
f(x)'=50x^49-10(6*ln5)(5^6x)
f(x)'=50x^49-60(ln5)(5^6x)

By Newton's Method
take Xo=1

Xi=1- f(Xo)/ f(Xo)'
=1-0.10356
=0.895644

Xii=1- f(Xi)/ f(Xi)'
=1-0.10356
=0.895644

so x=0.896 (correct to 3 sig. fig.)

1. I think you ask: 2^(2x-3)-3‧2^(x-2)+1=0
If yes,
Let y=2^(x-2)
then y^2=[2^(x-2)]^2=2^[2(x-2)]=2^(2x-4)=2^[(2x-3)-1]=[2^(2x-3)]/2
so 2^(2x-3)=2y^2

2y^2-3y+1=0
(2y-1)(y-1)=0
y=1/2 or 1
2^(x-2)=1/2 or 1
2^(x-2)=2^(-1) or 2^0
x-2=-1 or 0
x=1 or 2

2.???