SOLUTION: log(x+1)+log(x-1)^2=0

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Question 717594: log(x+1)+log(x-1)^2=0
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%28x%2B1%29%29%2Blog%28%28%28x-1%29%5E2%29%29=0
Solving logarithmic equations like this one usually starts with transforming the equation into one of the following forms:
log(expression) = number
or
log(expression) = log(expression)

With the "non-log" term of zero, your equation will be difficult to transform the equation into the "all-log" second form. So we will aim for the first form. This will require that we find a way to combine the two logs on the left side into one. Fortunately there is a property of logarithms,
We now have the desired form.

The next step with this form is to rewrite the equation in exponential form. In general, log%28a%2C+%28p%29%29+=+n is equivalent to p+=+a%5En. Using this pattern on our equation (and the fact that the base of "log" is 10) we get:
%28x%2B1%29%2A%28x-1%29%5E2=10%5E0
which simplifies as follows...
%28x%2B1%29%2A%28x-1%29%5E2=1
The (x+1) times the first (x-1):
%28x%5E2-1%29%2A%28x-1%29=1
Using FOIL:
x%5E3-x%5E2-x%2B1+=+1

With the x's out of the logarithms we can now solve for x. With a x%5E3 we are going to need one side to be zero (by subtracting 1):
x%5E3-x%5E2-x+=+0
and then we factor. First the greatest common factor of x:
x%28x%5E2-x-1%29+=+0
Nothing will factor any further. So we use the Zero Product Property:
x+=+0 or x%5E2-x-1+=+0
The first equation already gives us a solution for x. We can use the Quadratic Formula to find solutions to the second equation:
x+=+%28-%28-1%29+%2B-+sqrt%28%28-1%29%5E2-4%281%29%28-1%29%29%29%2F2%281%29
Simplifying...
x+=+%28-%28-1%29+%2B-+sqrt%281-4%281%29%28-1%29%29%29%2F2%281%29
x+=+%28-%28-1%29+%2B-+sqrt%285%29%29%2F2%281%29
x+=+%281+%2B-+sqrt%285%29%29%2F2
which is short for:
x+=+%281+%2B+sqrt%285%29%29%2F2 or x+=+%281+-+sqrt%285%29%29%2F2
These two, plus the x = 0 we found earlier, give us three possible solutions to our equation.

Solving logarithmic equations like this one requires that we do at least a partial check. For each possible solution we must at least make sure that every base and every argument of every logarithm is valid for that value of x. (Valid bases and arguments are positive and valid bases are not equal to 1.) Use the original equation to check:
log%28%28x%2B1%29%29%2Blog%28%28%28x-1%29%5E2%29%29=0

Checking x = 0:
log%28%28%280%29%2B1%29%29%2Blog%28%28%28%280%29-1%29%5E2%29%29=0
Simplifying...
log%28%281%29%29%2Blog%28%28%28-1%29%5E2%29%29=0
log%28%281%29%29%2Blog%28%281%29%29=0
At this point we can see that the bases are 10's and the arguments are 1's, all valid. So x = 0 passes the check.

Checking x+=+%281+%2B+sqrt%285%29%29%2F2:
For this we need a decimal approximation. Using a calculator we get (rounded to three places): 1.618
log%28%28%281.618%29%2B1%29%29%2Blog%28%28%28%281.618%29-1%29%5E2%29%29=0
Simplifying...
log%28%282.618%29%29%2Blog%28%28%280.618%29%5E2%29%29=0
At this point we can see that the bases are 10's and the arguments are (or will be after we square 0.618) positive, all valid. So x+=+%281+%2B+sqrt%285%29%29%2F2 passes the check.

Checking x+=+%281+-+sqrt%285%29%29%2F2:
For this we need a decimal approximation. Using a calculator we get (rounded to three places): -0.618
log%28%28%28-0.618%29%2B1%29%29%2Blog%28%28%28%28-0.618%29-1%29%5E2%29%29=0
Simplifying...
log%28%280.382%29%29%2Blog%28%28%28-1.618%29%5E2%29%29=0
At this point we can see that the bases are 10's and the arguments are (or will be after we square -1.618) positive, all valid. So x+=+%281+-+sqrt%285%29%29%2F2 passes the check.

So there are three solutions to your equation:
x+=+0 or x+=+%281+%2B+sqrt%285%29%29%2F2 or x+=+%281+-+sqrt%285%29%29%2F2