`e, 2^x . 16 = 4^x`
`⇒ 2^x . 2^4 = 4^x`
`⇒ 2^(x + 4) = (2^2)^x`
`⇒ 2^(x + 4) = 2^(2 . x)`
`⇒ x + 4 = 2 . x`
`⇒ 2 . x - x = 4`
`⇒ x = 4`
Vậy `x = 4`
`f, 5^(2x + 1) + 25^(x + 2) = 125 + 2020^0`
`⇒ 5^(2x + 1) + 5^(2x + 4) = 125 + 1`
`⇒ 5^(2x + 1) + 5^(2x + 1) . 5^3 = 126`
`⇒ 5^(2x + 1) . (1 + 5^3) = 126`
`⇒ 5^(2x + 1) . 126 = 126`
`⇒ 5^(2x + 1) = 1`
`⇒ 2x + 1 = 0`
`⇒ 2x = 0 - 1`
`⇒ 2x = -1`
`⇒ x = (-1)/2`
Vậy `x = (-1)/2`
