How do you solve #Log (x+5) - log (x-1) = log (x+2) - log (x-3)#?

Question

7165 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-21T15:01:13

#x=13#

To do this problem, you must know that #log(a)+log(b)=log(ab)# and that #log(a)-log(b)=log(a/b)#.

#log(x+5)-log(x-1)=log(x+2)-log(x-3)#

#log(x+5)+log(x-3)-log(x-1)-log(x+2)=0#

#log(((x+5)(x-3))/((x-1)(x+2)))=0#

Remember that #log(a)# is another way of writing #log_10(a)#.

#10^(log(((x+5)(x-3))/((x-1)(x+2))))=10^0#

#((x+5)(x-3))/((x-1)(x+2))=1#

#(x+5)(x-3)=(x-1)(x+2)#

#x^2+2x-15=x^2+x-2#

#2x-15=x-2#

#x=13#