Question

方程log2(9^(x-1)+7)=2+log2(3^(x-1)+1)的解?

Answer 1

log2[9^(x-1) +7]-log2[3^(x-1) +1]=2

即log2{[9^（x-1）+7]/[3^（x-1）+1]}=2

[9^（x-1）+7]/[3^（x-1）+1]=4

9^(x-1)+7=4*3^(x-1)+4

令3^(x-1)=t,则9^(x-1)=t^2

t^2-4t+3=0

(t-1)(t-3)=0

t=1,t=3

t=1时,3^(x-1)=1,则有x=1

t=3时,3^(x-1)=3,则x=2

代入检验,符合.