Verified answer

First, exponentiate both sides:

e^( ln(y) - ln(y-4) ) = e^( ln(2) )

Subtracting natural logs i like dividing the insides, and the natural log on the right side cancels out:

e^( ln( y / (y-4) ) = 2

Now that the natural log on the left side is one term, it can also cancel out:

y / (y-4) = 2

Solve for y:

y = 2(y - 4)

y = 2y - 8

-y = -8

y = 8

Recall the following logarithmic identity:

ln(a) + ln(b) = ln(ab)

ln(a) - ln(b) = ln(a / b)

Applying this identity, we can simplify the left side of the equation:

ln(y) - ln(y - 4) = ln(2)

ln(y / (y - 4)) = ln(2)

Next, recall that the definition of the natural logarithm states that the following two statements are equivalent:

e^a = x

ln(x) = a

A corollary of this definition is the following pair of identities:

ln(e^a) = a

e^ln(a) = a

This means that logarithmic and exponential functions are inverses. Accordingly, we can clear the logarithms from both sides of our equation by exponentiating:

ln(y / (y - 4)) = ln(2)

e^ln(y / (y - 4)) = e^ln(2)

y / (y - 4) = 2

This is a much simpler equation to solve. We simply gather like terms:

y = 2(y - 4)

y = 2y - 8

y = 8

The left side equals ln[ y/(y-4) ], by a property of logarithms. So this problem is equivalent to: y/(y-4) = 2, which I think you can solve.

Ln(a) - Ln(b) = Ln(a/b)..

Ln(y) - Ln(y-4) = Ln(2)

Ln(y/(y-4)) = Ln(2)

y/(y-4) = 2 { Raise both sides to the "e" }

y = 2y - 8

2y - y = 8

y = 8

ln(y) - ln(y-4) = ln(2)

ln( y / (y-4) ) = ln(2)

y / (y-4) = 2

y = 2y - 8

-y = -8

y = 8

first divide everything by ln and get

y-(y-4)= 2

4 doesn't equal 2

equations is not equal

there is no solution

ln [ y / (y - 4) ] = ln 2

y / (y - 4) = 2

y = 2y - 8

8 = y