How do you solve #log_3 x-log_3(x+5)=3#?

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Answer 1 · 2016-09-11T12:28:32

#x = -135/26 = 5 5/26#

"If the log terms are being subtracted, then the numbers are being divided"

#log_3 (x/(x+5) )= 3#

"Both sides must be logs, or both sides must be numbers"

#log_3 3 =1 " " rarr 3log_3 3 =3xx1 = 3#

#log_3 (x/(x+5) )= 3log_3 3#

#log_3 (x/(x+5) )= log_3 3^3#

#:. x/(x+5) =27/1 " "(3^3 = 27)#

#27(x+5) = x#

#27x +135 = x#

#26x = -135#

#x = -135/26#