Please help me with this logarithmic equation. How do I solve it?

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Answer 1 · 2018-05-28T12:52:02

Solution : $x=4$

$log_8 (x+4) + log_8 (x-3) = 1$ or

$log_8 {(x+4)(x-3)} =log_8 8 [log_b b=1]$ or

$(x+4)(x-3) = 8 or x^2 +x -12 -8 =0$ or

$x^2 +x -20 =0 or (x+5)(x-4)=0$ or

$x= -5 , x =4$ When $x= -5 ; x+4= -1$ (negative)

$:.x=-5$ is invalid . Solution is $x=4$ [Ans]

Answer 2 · 2018-05-28T12:52:35

$x=4$

Writing
$log_8(x+4)(x-3)=log_8 8$
so
$(x+4)(x-3)=8$
$x^2+4x-3x-12=8$
thus
$x^2+x-20=0$
with the Solutions
$x_1=4$
$x_2=-5$ (which is not a solution!)