Question

How do you solve log8x+log8(x+6)=log8(5x+12)?

Answer 1

$log_{8}(x) + log_{8}(x + 6) = log_{8}(5x + 12)$

$= log_{8}(x(x + 6)) = log_{8}(5x + 12)$

$log_{8}( {x}^{2} + 6x ) = log_{8}(5x + 12)$

${x}^{2} + 6x = 5x + 12$

${x}^{2} + x - 12 = 0$

${x}^{2} + 4x - 3x - 12 = 0$

$x(x + 4) - 3(x + 4) = 0$

$(x - 3)(x + 4) = 0$

So x= 3 or x= -4