SOLUTION: Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this. ln (x - 5) + ln (x + 4) = ln 36 So I thought I remembered how t

Question 212873: Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this.
ln (x - 5) + ln (x + 4) = ln 36
So I thought I remembered how to do these but I guess not. I tried to google some examples but still can't figure it out. Can someone help it would be greatly appreciated.
Found 2 solutions by algebrapro18, jsmallt9:
Answer by algebrapro18(249) (Show Source): You can put this solution on YOUR website!
Well to do this problem we need to use 2 properties of ln.

1.
2.

Now that we have those out of the way we can solve the equation. First we need to use property 1 on the equation. We get:

Now we use propterty 2 by raising both sides to the e power.

And now we solve the quadratic equation.

(x-5)(x+4) = 36
x^2-x-20 = 36
x^2-x-56 = 0
(x-8)(x+7)=0

so x = -7 and x = 8. If you check those solutions you will find that indeed they work.
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
Often the easiest way to solve equations with multiple logarithms is to start by combining logarithms. There are a couple of properties of logarithms we can use to do this:

We can use the first one on your equation:

The easiest way to proceed is to recognize that if , then .
The longer way is to:
Subtract ln(36) from both sides: ln((x-5)(x+4)) - ln(36) = 0
Use the second property above to combine the ln's on the left: ln((x-5)(x+4)/36) = 0
Rewrite the equation in exponential form:
Substitute 1 for (since any non-zero number to the zero power is 1) and multiply both sides by 36 leaving you with the equation we got in one step earlier: (x-5)(x+4) = 36

Either way we can now solve:

Multiply out the left side:

As a quadratic equation we should get one side equal to zero (by subtracting 36 from each side):

Now we can factor this (or use the quadratic formula):

In order for this product to be zero one fo the factors must be zero:
x - 8 = 0 or x + 7 = 0
Solving these we get:
x = 8 or x = -7
This is the solution to the quadratic equation. But since the arguments of logarithm functions, no matter what the base, must be positive, we must check these answers to make sure they will work in the original equation:
When x = 8, both of the ln's on the left have positive arguments so 8 works. But when x = -7, the ln's have negative arguments which we cannot allow. So we must reject x = -7. (Even if only one of the ln's ended up with a negative argument we would still have to reject -7!)
So our final answer is just x = 8. (You can check with a calculator to see that ln(8-5) + ln(8+4) = ln(36).)

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