Solve this equation please

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Kamen Minkov asked Apr 23, 2022

247,286 views

12 votes

Solve this equation please

Mathematics
high-school

Kamen Minkov asked Apr 23, 2022

by Kamen Minkov

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1 Answer

14 votes

Explanation:

Number 1

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

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