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

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Answer 1 · 2015-09-06T17:02:34

#x=4#

First we simplify the left side by using the rule: #log (a) - log(b) = log(a/b)#

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

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

Then, we apply the rule: #log(a) = b <=> 10^b = a#

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

Then, we can apply the rule: #10^log(k) = k#

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

Now we can just solve the equation:

#3x-6 = x+2#

#2x=8#

#x=4#

Alternatively, we can skip a lot of steps by taking a shortcut:

Since #log()# is a one-to-one function, #log(a) = log(b)# means #a=b#.

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

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

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

#x+2=3x-6#

#8=2x#

#4=x#