Solve the equation. Check for extraneous solutions.
ln x + ln(x-2) = 1
lnx + ln(x-2) = 1 , where x>2
ln(x(x-2)) = 1
x^2 - 2x = e^1
x^2 - 2x + 1 = e+1 , ( I completed the square)
(x-1)^2 = e+1
x - 1 = ± √(e+1)
x = 1 ± √(e+1)
but x > 2 , so
x = 1 + √(e+1)
To solve the equation ln(x) + ln(x-2) = 1, we can use the logarithmic properties. Firstly, we can apply the property of logarithms that states the sum of logarithms is equal to the logarithm of the product. Therefore, we can rewrite the equation as ln(x(x-2)) = 1.
Next, we can use the property that the logarithm and exponential functions are inverses of each other. By taking the exponential of both sides of the equation, the exponential function e^x cancels out the natural logarithm function ln(x(x-2)). Therefore, we can rewrite the equation as e^1 = x(x-2).
Simplifying the expression on the right-hand side, we get e = x^2 - 2x. Rearranging the equation to the standard form, we have x^2 - 2x - e = 0.
Now, we have a quadratic equation. In order to find the solutions, we can either factorize the equation or use the quadratic formula.
Factoring: Since the coefficient of x^2 is 1, the factors of the constant term (e) have a sum of -2. Factoring the quadratic equation, we have (x - 1)(x - e) = 0. Therefore, the solutions are x = 1 and x = e.
Using the quadratic formula: The quadratic formula is given by x = (-b ± sqrt(b^2 - 4ac))/(2a). In our case, a = 1, b = -2, and c = -e. Substituting these values into the quadratic formula, we have x = (-(-2) ± sqrt((-2)^2 - 4(1)(-e)))/(2(1)). Simplifying further, we get x = (2 ± sqrt(4 + 4e))/(2). Canceling out the common factors, we have x = 1 ± sqrt(1 + e).
Now that we have found the solutions, we need to check for extraneous solutions. In this case, we have an equation with a logarithm, and logarithms are only defined for positive values. Thus, we need to make sure that the solutions we found for x are positive.
Checking the solutions:
1. For x = 1:
ln(1) + ln(1-2) = 0 + ln(-1) = undefined. Therefore, x = 1 is not a valid solution.
2. For x = e:
ln(e) + ln(e-2) = 1 + ln(e-2) = 1 + ln(e-2) = undefined. Therefore, x = e is not a valid solution.
In conclusion, there are no valid solutions for the equation ln(x) + ln(x - 2) = 1, and thus, there are no extraneous solutions either.