Determine the coordinates of the point on the curve representing the function f(x)=ln(2x-1) where the tangent is parallel to the line with equation 2x-5y=0
Thanks for any help!
The slope of the 2x -5y = 0 line is 2/5.
You are looking for a point on the f(x) curve where f'(x) = 2/5
f'(x) = 2/(2x -1) = 2/5
x = 3
f(x) = ln(5) = 1.60944
To determine the coordinates of the point on the curve where the tangent is parallel to the line with equation 2x - 5y = 0, there are a few steps we need to follow:
Step 1: Find the derivative of the function f(x) = ln(2x - 1).
The derivative of ln(u) with respect to x is given by f'(x) = (1/u) * u', where u' is the derivative of the function inside the natural logarithm. In this case, the derivative of (2x - 1) is simply 2.
Therefore, f'(x) = (1/(2x - 1)) * 2 = 2/(2x - 1).
Step 2: Determine the slope of the tangent line parallel to the line with equation 2x - 5y = 0.
The given equation of the line is in the form of y = (2/5)x, where the coefficient of x represents the slope of the line. So, the slope of the line is 2/5.
Since we want the tangent line to be parallel to this line, the slope of the tangent line should also be 2/5.
Step 3: Set the derivative equal to the slope of the tangent line.
Set the derivative of the function, f'(x), equal to the slope of the tangent line, which is 2/5.
2/(2x - 1) = 2/5.
Step 4: Solve for x.
Cross-multiply and simplify the equation:
5 * 2 = 2 * (2x - 1),
10 = 4x - 2.
4x = 12,
x = 3.
Step 5: Substitute the value of x into the original function to find y.
Substitute x = 3 into the original function f(x) = ln(2x - 1):
f(3) = ln(2 * 3 - 1) = ln(5) ≈ 1.6094.
Therefore, the point on the curve representing the function f(x) = ln(2x - 1) where the tangent is parallel to the line 2x - 5y = 0 is (3, ln(5)) or approximately (3, 1.6094).