SOLUTION: find the solution to the logarithmic equation ln(x+6)+ln(x-6)=0 i tried cancelling the ln by multiplying by e i get (x+6)+(x-6)=0 2x=0 x=0 but this is not right..what s

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: find the solution to the logarithmic equation ln(x+6)+ln(x-6)=0 i tried cancelling the ln by multiplying by e i get (x+6)+(x-6)=0 2x=0 x=0 but this is not right..what s      Log On


   

Question 267953: find the solution to the logarithmic equation
ln(x+6)+ln(x-6)=0
i tried cancelling the ln by multiplying by e
i get (x+6)+(x-6)=0
2x=0
x=0
but this is not right..what should i do?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
ln%28x%2B6%29%2Bln%28x-6%29=0

I'm glad you found that multiplying by e didn't work. Try to forget the idea that e*ln cancels out ASAP!!

Solving equations where the variable is in the argument of a logarithm, like your equation, usually starts by transforming the equation into one of the following forms:
log(variable-expression) = other-expression
or
log(variable-expression) = log(other-expression)

With the zero term on the right side, it will be difficult to get an "all logarithm" equation like the second form. So we will aim for the first form. This will require us to combine the two logarithms into one.

Since logarithms can be added only if both the bases and the arguments are the same. Your logarithms have the same base but the arguments are different. So we cannot add them together. But there is another way to combine the two. There is a property, log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29, which does provide a way to combine two logarithms of the same base with a plus sign between them. Using this on your equation we get:
ln%28%28x%2B6%29%2A%28x-6%29%29=0
which simplifies to:
ln%28x%5E2-36%29=0
We now have the equation in the desired form. With the first form we proceed by rewriting the equation in exponential form:
x%5E2+-+36+=+e%5E0
which simplifies to:
x%5E2+-+36+=+1
We now have a quadratic equation to solve. So we get one side equal to zero (by subtracting 1 from each side):
x%5E2+-+37+=+0
And then we try to factor this or use the Quadratic formula. The left side does not factor so we must use the Quadratic formula. You should find that:
x+=+sqrt%2837%29 or x+=+-sqrt%2837%29

When solving logarithmic equations you should always check your answers. One important thing to check is the arguments of the logarithms. They should never be zero or negative.

ln%28x%2B6%29%2Bln%28x-6%29=0
Checking x+=+sqrt%2837%29:
ln%28%28sqrt%2837%29%29%2B6%29%2Bln%28%28sqrt%2837%29%29-6%29=0
Since sqrt%2837%29+%3E+6 both arguments are positive. This solution, if you finish the check, works.

Checking x+=+-sqrt%2837%29:
ln%28%28-sqrt%2837%29%29%2B6%29%2Bln%28%28-sqrt%2837%29%29-6%29=0
Since sqrt%2837%29+%3E+6 both arguments are negative. Therefore this solution must be rejected. (Note: If even only one argument was zero or negative you would still reject this solution.)

So the only solution to the equation is x+=+sqrt%2837%29