How do you solve #ln2-ln(3x+2)=1#?

1 Answer
Jul 13, 2016

Answer:

#x = (2(1/e-1))/3 ≈ -0.42141#

Explanation:

In order to solve this logarithmic equation, we can make use of the properties of logarithms, such as

Property:
#ln(a) - ln(b) = ln(a/b)#

We can now rewrite this equation as follows:

#ln(2/(3x+2)) = 1#

To get rid of the natural logarithm on the left-hand side, we take the #e#-xponential on both sides, giving us

#2/(3x+2) = e^1#

To simplify this even further and solve for #x#, the best thing to do here would be to to get rid of the fraction. We can do this by multiplying both sides by #3x+2#, which yields

#2/(cancel((3x+2)))*cancel((3x+2)) = e(3x+2) #

So our equation now has become a lot more appealing:

#2 = e(3x+2)#

Since #e# is just a constant, namely #e ≈ 2.718#, we can divide both sides #e# to isolate the term with the #x# in it.

#2/e = 3x+2#

Subtracting #2# from both sides gives us

#2/e - 2 = 3x#

Dividing then by #3# yields

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

Or we can factor out a #2# and write:

#x = (2(1/e-1))/3 ≈ -0.42141#