Verified answer

log_3(x^2 − 3x − 31) = 2

x² - 3x - 31 = 9

x² - 3x - 40 = 0

(x - 8)(x + 5) = 0

x = 8 or x = -5

log(x) = log(2) + log(4x − 6)

log(x) = log(8x − 12) ← product rule

x = 8x − 12

12 = 7x

x = 12/7

♣♦

log(base 2) (2x-4) = 2 + log(base 2) (x^2-6) => log(base 2) (2x-4) - log(base 2) (x^2-6) = 2 => log(base 2) [(2x - 4) / (x^2 - 6) ] = 2 => (2x - 4) /(x^2 - 6) = 2^2 = 4 => 2x - 4 = 4x^2 - 24 => divide by 2 x - 2 = 2x^2 - 12 => 2x^2 - x - 10 = 0 => 2x^2 - 5x + 4x - 10 = 0 => x(2x - 5) + 2(2x - 5) = 0 => (2x - 5)(x + 2) = 0 since log of negative numbers is not defined 2x = 5 x = 5/2