a,

`2^(x-2)` = `1/8` = `2^(-3)`

x-2 = -3

x = -1

b,

`5^(x^2-8x+12)` = 1 = `5^0`

`x^2-8x+12` = 0

(x-6)(x-2) = 0

`x_1` = 6

`x_2` = 2

c,

`2*3^(x+3)` - `5*3^(x-2)` = 1443

`2*3^3*3^x - 5*(1/9)*3^x` = 1443

`(54-5/9)*3^x` = 1443

`3^x` = 27 = `3^3`

x = 3

d,

`7^(x-3)` = `4^(2x-6)` = `16^(x-3)`

Különböző alapok, azonos kitevők, csak akkor lehet, ha a kitevő 0.

x-3 = 0

x = 3

e,

`4^x-9*2^x+8` = 0

`2^(2x)-9*2^x` + 8 = 0

`2^x` = a helyettesítéssel.

`a^2-9a+8` = 0

(a-1)(a-8) = 0

`a_1` = 1 = `2^x`

`x_1` = 0

`a_2` = 8 = `2^x`

`x_2` = 3