SOLUTION: Find the exact solution, using common logarithms. log(x^(2)+3)-log(x+3)=2+log(x-3)

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Question 626908: Find the exact solution, using common logarithms.
log(x^(2)+3)-log(x+3)=2+log(x-3)

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%28x%5E%282%29%2B3%29%29-log%28%28x%2B3%29%29=2%2Blog%28%28x-3%29%29
Solving logarithmic equations like this usually starts with transforming the equation, with algebra and/or properties of logarithms, into one of the following forms:
log(expression) = other-expression
or
log(expression) = log(other-expression)

Since your equation has a "non-log" term, the 2, it will be more difficult to achieve the second, "all-log", loorm. So we will aim for the first form.

Your equation has three log terms and we want only one. Somehow we need to find a way to eliminate and/or combine these logarithms into one. First we will gather them all on the left side by subtracting the log on the right side from both sides of the equation:
log%28%28x%5E%282%29%2B3%29%29-log%28%28x%2B3%29%29-log%28%28x-3%29%29=2

None of these log terms are like terms. So we cannot add or subtract them (which would have been a way to combine/eliminate them). Fortunately there are properties of logarithms that provide an alternate way of combining logs:
  • log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Aq%29%29
  • log%28a%2C+%28p%29%29+-+log%28a%2C+%28q%29%29+=+log%28a%2C+%28p%2Fq%29%29
These properties require that the two logs have the same base and have coefficients of one. All three of your logs meet these requirements so we can use these properties. Since we have two subtractions, we will use the second property above. Using it on the first two logs we get:
log%28%28%28x%5E%282%29%2B3%29%2F%28x%2B3%29%29%29-log%28%28x-3%29%29=2
Using the property again on the remaining logs we get:
log%28%28%28%28x%5E%282%29%2B3%29%2F%28x%2B3%29%29%2F%28x-3%29%29%29=2
We now have the first form.

The next step with the first form is to rewrite the equation in exponential form. In general log%28a%2C+%28p%29%29+=+q is equivalent to a%5Eq+=+p. Using this pattern (and the fact that the base of log is 10) on our equation we get:
10%5E2+=+%28%28x%5E%282%29%2B3%29%2F%28x%2B3%29%29%2F%28x-3%29
which simplifies to:
100+=+%28%28x%5E%282%29%2B3%29%2F%28x%2B3%29%29%2F%28x-3%29

Now we solve for x. First we eliminate the fraction(s). Muliplying both sides by x-3:
100x-300+=+%28x%5E%282%29%2B3%29%2F%28x%2B3%29
Multiplying both sides by x+3 (using FOIL on the left side) we get:
100x%5E2-900+=+x%5E2%2B3
Since there is just x%5E2 terms and no "x" terms, we can solve this without factoring or the Quadratic Formula. We can use gather the squared terms on one side and the numbers on the other. Subtracting x%5E2 and adding 900 we get:
99x%5E2+=+903
Dividing both sides by 99 we get:
x%5E2+=+903%2F99
Since 99 and 903 are both divisble by 3 we can reduce the fraction:
x%5E2+=+301%2F33
Now we find the square root of each side (remembering that there is a positive and a negative square root)
x+=+sqrt%28301%2F33%29 or x+=+-sqrt%28301%2F33%29
Next we rationalize the denominators:
x+=+sqrt%28%28301%2F33%29%2833%2F33%29%29 or x+=+-sqrt%28%28301%2F33%29%2833%2F33%29%29
x+=+sqrt%289933%2F33%5E2%29 or x+=+-sqrt%289933%2F33%5E2%29
x+=+sqrt%289933%29%2Fsqrt%2833%5E2%29 or x+=+-sqrt%289933%29%2Fsqrt%2833%5E2%29
x+=+sqrt%289933%29%2F33 or x+=+-sqrt%289933%29%2F33

Last of all we check our solutions. This is not optional! You must at least check to see that all arguments to all logarithms remain positive. Use the original equation to check:
log%28%28x%5E%282%29%2B3%29%29-log%28%28x%2B3%29%29=2%2Blog%28%28x-3%29%29
Checking x+=+sqrt%289933%29%2F33:

The first two logs, with multiplication and addition of positive numbers, will obviously turn our positive. But what about the third log. What will be get if we subtract 3 from the x value? The quickest way to find out is to your your calculator to get a decimal approximation for sqrt%289933%29%2F33%29. You should find that it is approximately 3.02013445. Subtracting 3 from this will result in 0.02013445, a very small but positive number. So all three arguments are positive. This solution passes the required check.

Checking x+=+-sqrt%289933%29%2F33:

We don't have to look any further than the third log. A negative number, like the x value, minus 3 will be negative. It doesn't matter what happens in the other logs. This solution fails this check. So we reject/discard it. (NOTE: This solution failed not because the x was negative but because it made a log's argument negative. There will be times when negative x's pass this check. And there will be times when non-negative x's will fail. You just have to see what happens in each argument to determine whether the solution checks out or not.)

So the only (exact) solution to this equation is:
x+=+sqrt%289933%29%2F33