How do you solve for x in # ln(x+2) - ln(x-2) = ln3#?

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Answer 1 · 2016-04-16T08:52:04

#x = 4#

One #log# minus another #log# can be condensed as the inside of the first divided by the inside of the second, all inside a single #log#, so

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

which gives us

#ln((x+2)/(x-2)) = ln3#.

Raising both sides by #e# and using simple algebraic manipulation, we can solve for #x#,

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

#x+2 = 3(x-2)#
#x+2 = 3x - 6#
#8 = 2x#
#4 = x#