How do you solve $log_19(-5x-6)=log_19(2-3x)$ ?

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Answer 1 · 2016-07-15T09:10:58

$x=-4$

$log_19(-5x-6)=log_19(2-3x)$

$hArrlog_19(-5x-6)-log_19(2-3x)=0$ or

$log_19((-5x-6)/(2-3x))=log_19(1)$ or

$(-5x-6)/(2-3x)=1$ or

$-5x-6=2-3x$ or

$-5x+3x=2+6$ or $-2x=8$

or $x=-8/2=-4$

Answer 2 · 2016-07-15T09:29:31

$x = -4$

The log equation leads directly to a linear equation.

if $logA = logB, " " rArr A = B$

$-5x -6 = 2 - 3x$

$-6-2 =-3x+5x$

$-8 = 2x$

$-4 = x$

How do you solve log_19(-5x-6)=log_19(2-3x)log_19(-5x-6)=log_19(2-3x)?