How do you solve #log_5 (x+3)= 3 + log_5 (x-3) #?

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Answer 1 · 2016-04-10T09:51:47

#x=189/62=3 3/62#

As #log_ba=loga/logb#

#log_5(x+3)=3+log_5(x-3)# can be written as

#log(x+3)/log5=3+log(x-3)/log5#

As #log5!=0#, multiplying both sides by #log5#

#log(x+3)=3log5+log(x-3)# or

#(x+3)=5^3(x-3)=125(x-3)# or

#x+3=125x-375# or

#124x=378# and #x=378/124=189/62=3 3/62#