Solve for x log3x 2 log3x 1 log32Solutionlog3 x2 log3 x

Solve for x: log3(x - 2) + log3(x - 1) = log32

Solution

log₃ (x-2) + log₃ (x-1) = log₃ 2

now log ab = log a + log b

so log₃ (x-2) + log₃ (x-1) =   log₃ (x-2)*(x-1)

log₃ (x-2)*(x-1) =  log₃ 2

(x-2) * (x-1) =2

(x^2 -3x +2) = 2

x^2 -3x +2 =2

x(x-3) =0

so x= 3 or x =0