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

1 Answer
Jul 7, 2016

Answer:

Using the properties of the logarithm we find that #x~=4.91#

Explanation:

To solve this equation we need to perform the "anit-log" of the equation, which is the exponential function or #e^x# where

#e^[ln(x)]=x#

However, there are two terms in the left-hand-side which are added together making our job more difficult. To get around this, we can use properties of the log function to make things simpler, in this case we use:

#ln(a*b)=ln(a)+ln(b)#

applied to our equation we get:

#ln[(x+2)(x-2)]=3#

Now we can take the exponential of both sides to get

#(x+2)(x-2)=e^3#

expanding the left-hand-side we get:

#x^2-4=e^3#

#x^2 = e^3+4#

#x=+-sqrt(e^3+4)#

#x~=+-4.91#

However, since we can't take the log of a negative number we know that the negative solution is spurious since both of our terms in the original equation would be negative inside the #ln# function. Therefore the solution is:

#x=sqrt(e^3+4) ~= 4.91#