What are the possible values of x for..?

Question

What are the possible values of x for:

1. (3logx)^2 + logx^6 - 8 = 0

2. e^x - 6e^-x > 1

Answer 1

(3logx)^2 + logx^6 - 8 = 0

9log^2 (x)+6log x- 8=0

log x=y

9y^2+6y-8=0

(3y+1)^2=9

y=2/3 or y = -4/3

then

x=10^(2/3) or x= 10^(-4/3)

e^x-6e^(-x)>1

e^x>0 for any x muliply by e^x and obtain

(e^x)^2-e^x-6>0

e^x = y

y^2-y-6>0

(y-3)(y+2)>0

y<-2 or y>3

solve for x

e^x<-2 no real x

then

e^x>3

x>ln3