How would I go about solving for x in the equation:
ln(4x-2)-ln(4)=-ln(x-2)
Thanks in advance
ln[(4x-2)/4] = ln[1/(x-2)]
(4x -2)/4 = 1/(x-2)
x - 1/2 = 1/(x-2)
x^2 -x/2 -2x +1 = 1
x^2 -(5/2)x = 0
x = 0 or 5/2
0 is not allowed since ln of a negative number is undefined
x = 5/2
To solve for x in the equation ln(4x-2) - ln(4) = -ln(x-2), you can follow these steps:
Step 1: Combine the logarithms on the left side of the equation using logarithmic properties.
ln((4x-2)/4) = -ln(x-2)
Step 2: Use the property of logarithms that states ln(a) = -ln(b) is equivalent to ln(a) + ln(b) = 0.
ln((4x-2)/4) + ln(x-2) = 0
Step 3: Combine the logarithms on the left side using the property ln(a) + ln(b) = ln(a * b).
ln(((4x-2)/4) * (x-2)) = 0
Step 4: Set the expression inside the logarithm equal to 1 since ln(1) = 0.
((4x-2)/4) * (x-2) = 1
Step 5: Simplify and solve the resulting equation.
Multiply both sides of the equation by 4 to eliminate the fraction:
(4x-2) * (x-2) = 4
Expand and simplify the left side:
4x^2 - 12x + 8 = 4
Rearrange the equation to form a quadratic equation:
4x^2 - 12x + 4 = 0
Step 6: Solve the quadratic equation. You can use factoring, the quadratic formula, or completing the square to find the solutions.
Factoring:
Divide each term by 4:
x^2 - 3x + 1 = 0
The quadratic equation does not factor easily, so you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -3, and c = 1, so the quadratic formula becomes:
x = (-(-3) ± √((-3)^2 - 4(1)(1))) / (2(1))
x = (3 ± √(9 - 4)) / 2
x = (3 ± √5) / 2
Therefore, the solutions to the equation ln(4x-2) - ln(4) = -ln(x-2) are:
x = (3 + √5) / 2 and x = (3 - √5) / 2.