Question

Solve this equation please​

Answer 1

Explanation:

Number 1

`log_(10)( {x}^(2) - 3x + 12 ) = 1 \\ log_(10)( {x}^(2) - 3x + 12 ) = log_(10)(10) \\ cancelling \: both \: logs \: we \: have \\ {x}^(2) - 3x + 12 = 10 \\ {x}^(2) - 3x + 12 - 10 = 0 \\ {x}^(2) - 3x + 2 = 0 \\`

`solve \: \: by \: factorisation \: method \\ {x}^(2) - 2x - x + 2 \\ {x}(x - 2) \: - 1(x - 2) \\ (x - 2)(x - 1) \\ therefore \: x = 2 \: or \: x = 1`

Number 2