Question

I tried to do these questions but I can't figure it out.
1)4^x+6(4^-x)=5
Textbook Answer=0.79,0.5

2)8(5^2x)+8(5^x)=6
Textbook Answer=-0.43

Answer 1

4^x+6(4^-x)=5
4^x + 6/4^x = 5
let 4^x = y , so you have
y + 6/y = 5
y^2 + 6 = 5y
y^2 - 5y + 6 = 0
(y - 2)(y - 3) = 0
y = 2 or y = 3

case1: 4^x = 2
(2^2)^x= 2^1
2^(2x) = 2^1
2x = 1
x = 1/2

case2: 4^x = 3
take log of both sides, and follow through with log rules
log(4^x) = log3
x log4 = log3
x = log3/log4 = appr .79 , same as the given answers

#2 8(5^2x)+8(5^x)=6
same way, let y = 5^x to get
8y^2 + 8y - 6= 0
4y^2 + 4y - 3 = 0

carry on as above, it will factor
by the looks of things, one answer will not be valid