How do you solve #log(x+3)-log(x-3)=2#?

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Answer 1 · 2015-10-31T11:06:54

#x=101/33#

Use the identity #log(a)-log(b)=log(a/b)#.

#log(x+3)-log(x-3)=2#

#log(frac{x+3}{x-3})=2#

#frac{x+3}{x-3}=10^2=100#

#x+3=100(x-3)=100x-300#

#303=99x#

#x=303/99=101/33#