SOLUTION: Solve the logarithmic equation. Be sure to reject any value that is not in the doman of the original logarithmic expression. Give the exact answer. ln(x-6)+ln(x+1)=ln(x-15)

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Question 277387: Solve the logarithmic equation. Be sure to reject any value that is not in the doman of the original logarithmic expression. Give the exact answer.
ln(x-6)+ln(x+1)=ln(x-15)
Answer by jsmallt9(3758) (Show Source):
You can put this solution on YOUR website!
Solving a logarithmic equation often starts by transforming the equation into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

Since your equation has nothing but logarithms, we'll aim for the second form. This will require that we combine the two logarithms on the left into one. They are not like terms so we cannot use regular addition. But there is a property of logarithms, , which allows you to combine two logarithms of the same base with a "+" between them. Using this on the two logarithms on the left we get:
ln((x-6)*(x+1)) = ln(x-15)
We now have the desired form. With this form we have two logarithms that are equal. And if the logarithms are equal then the arguments must be equal:
(x-6)*(x+1) = x-15
Now we have an equation we can solve. Multiplying out the left side we get:

This is a quadratic equation so we want one side to be zero. Subtracting x from and adding 15 to each side we get:

Now we factor (or use the Quadratic Formula). This factors easily:

From the Zero Product Property we know that this product is zero only if one of the factors is zero. Since both factors are the same, x-3 must be zero. Solving for x we get
x = 3

When solving logarithmic equations it is important (not just a good idea) to check your answers. And use the original equation when checking:
ln(x-6)+ln(x+1)=ln(x-15)
Checking x = 3:
ln((3)-6)+ln((3)+1)=ln((3)-15)
which simplifies to:
ln(-3)+ln(4)=ln(-12)
As you can see, two of the arguments are negative. Arguments of logarithms must always be positive. They can never be zero or negative. Values of x make an argument zero or negative are not allowed to be in the domain. So x=3 is not in the domain. We must reject this solution.

Since x=3 was the only possible solution and since we had to reject it, there is no solution for your equation. (IOW, the original equation was impossible to begin with.)