1. 4^2x = 32^1/2
2. 9^x = 3^x-2
3. 2^x = 4^x+1
4. 4^x = 10
5. (1/4)^2x = (1/2)^x
6. 2.4^3x+1 = 9
1. 4^2x = 32^1/2 ......the idea is to get the bases the same and then solve for the exponents
(2^2) ^(2x) = (2^5)^(1/2)
(2)^(4x) = (2)^(5/2)
4x = 5/2 divide both sides by 4
x = 5/8
2. 9^x = 3^x-2
(3^2)^x =3^(x - 2)
(3)^(2x) = 3^(x - 2)
2x = x - 2 subtract x from both sides
x = - 2
3. 2^x = 4^x+1
2^x = (2^2)^(x + 1)
2^x = (2)^(2x + 2)
x = 2x + 2 subtract 2x from both sides
-1x = 2 divide both sides by -1
x = -2
4. 4^x = 10
Take the log of each side
log 4^x = log 10 and we can write
x * log 4 = 1 divide both sides by log 4
x = 1 / log 4
5. (1/4)^2x = (1/2)^x
[ (1/2)^2 ]^(2x) = (1/2)^x
(1/2)^(4x) = (1/2)^x
4x = x
4x - x = 0
3x = 0
x = 0