Question

7. Solve forse, log (x-1) +
log (x- 2) = log 2 +log3​

Answer 1

x = - 1, 4

Explanation:

`log(x - 1) + log(x - 2) = log \: 2 + log \: 3 \\ \\ log \{(x - 1) (x - 2) \} = log \{2 * 3 \}\\ \\ log \{ {x}^(2) - 2x - x + 2 \} = log \:6\\ \\ {x}^(2) - 3x+ 2 = 6 \\ \\ {x}^(2) - 3x+ 2 - 6 = 0\\ \\ {x}^(2) - 3x - 4 = 0\\ \\ {x}^(2) - 4x + x - 4 = 0\\ \\ x(x - 4) + 1(x - 4) = 0 \\ \\ (x - 4)(x + 1) = 0 \\ \\ x - 4 = 0 \: \: or \: \: x + 1 = 0 \\ \\ x = 4 \: \: or \: \: x = - 1 \\ \\ x = - 1, \: \: 4`