How do you solve $log_3x+log_3(x-8)=2$ ?

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Answer 1 · 2018-03-31T13:54:27

Thus, x=9 is the valid solution

$log_3 x+log_3 (x-8)=2$

$logm+logn=logmn$

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

$log_3 x(x-8)=2$

$x(x-8)=3^2$

$x(x-8)=9$

$x^2-8x=9$

$x^2-8x-9=0$

$x^2-9x+1x-9=0$

$x(x-9)+1(x-9)=0$

$(x+1)(x-9)=0$

$x=-1, x=9$

$x=-1,9$

Here, x needs to be a positive number .

Thus, x=9 is the valid solution

How do you solve log_3x+log_3(x-8)=2log_3x+log_3(x-8)=2?