At what point to the curve y=4+3lnx is the tangent line paralell to the line 5y-x=10.
3/x = 1/5 --->
x = 15
To find the point where the tangent line to the curve y = 4 + 3ln(x) is parallel to the line 5y - x = 10, we need to find the x-coordinate of the point where the tangent line has the same slope as the line 5y - x = 10.
The slope of the line 5y - x = 10 can be found by rearranging the equation in slope-intercept form (y = mx + b), where m is the slope:
5y - x = 10
5y = x + 10
y = (1/5)x + 2
So, the slope of the line is 1/5.
Now, we need to find the derivative of the curve y = 4 + 3ln(x) to find its slope at any given point.
To find the derivative, we use the chain rule:
dy/dx = d(4 + 3ln(x))/dx
= 0 + 3(1/x)
= 3/x
To find the x-coordinate where the tangent line has the same slope as the line 5y - x = 10, we need to set the derivative of the curve equal to the slope of the line:
3/x = 1/5
To solve for x, we can cross-multiply:
5 * 3 = 1 * x
15 = x
So, the x-coordinate where the tangent line is parallel to the line 5y - x = 10 is x = 15.
Therefore, the point where the tangent line is parallel to the line is (15, y).
To find the corresponding y-coordinate, substitute x = 15 into the equation y = 4 + 3ln(x):
y = 4 + 3ln(15)
≈ 10.20
So, the point where the tangent line is parallel to the line 5y - x = 10 is approximately (15, 10.20).