How do you solve $Log_3x + log_3(x-8) = log_3(8x)$ ?

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Answer 1 · 2016-03-22T20:53:19

$x=16$

$log_3 x+log_3 (x-8)-log_3(8x)=0$

$log_3((x(x-8))/(8x))=0$ -> use properties $log_b(xy)=log_bx+log_by, and log_b(x/y) = log_b x-log_b y$

$log_3 ((x-8)/8)=0$

$3^0=(x-8)/8$ -> use property $y=log_bx iff b^y=x$

$1=(x-8)/8$

$8=x-8$

$16=x$

How do you solve Log_3x + log_3(x-8) = log_3(8x)Log_3x + log_3(x-8) = log_3(8x)?