More calculus. . .?

Question

For what value of x does the tangent line to y = ln(3x) at x=2 cross the x-axis?

Edit: Please indicate how you got your answer so I can understand.

Answer 1

First, let's solve for the tangent line at x = 2.

y = ln(3x).

We can solve this in two ways. I'll show you the second way later.

y' = 1/(3x) [3] = 3/(3x) = 1/x

Therefore, we obtain our slope m when x = 4. That is
m = 1/4
When x = 4, y = ln(3*4) = ln(12).

We now use the slope formula with (4, ln(12)) and (x, y).

(y2 - y1) / (x2 - x1) = m
[y - ln(12)] / [x - 4] = 1/4. Cross multiplying,
4[y - ln(12)] = x - 4. Dividing both sides by 4, we have
y - ln(12) = (1/4)x - 1. Adding ln(12) both sides,

y = (1/4)x - 1 + ln(12)

That is the equation of our line. To determine where this crosses the x-axis, we find the x-intercept by making y = 0.

0 = (1/4)x - 1 + ln(12)

1 - ln(12) = (1/4)x, and then multiplying both sides by 4 gives
4 - 4ln(12) = x

Therefore, the tangent value to y = ln(3x) at x = 2 crosses the x-axis at x = 4 - 4ln(12).

****
The second method to solve for the derivative of y = ln(3x) is as follows.

y = ln(3x). Use the log property of addition.
y = ln(3) + ln(x)

NOW, when you take the derivative of both sides, ln(3) is a constant, so

y' = 1/x.

Just an FYI.

Answer 2

First find the slope of the tangent line by taking the derivative of the curve.

y = ln 3x
e^y = e^(ln 3x) = 3x
(e^y)(dy/dx) = 3
dy/dx = 3/e^y = 3/(3x) = 1/x

at x = 2
dy/dx = 1/x = 1/2

Plugging in

y = ln 3x = ln (3*2) = ln 6

The equation of the tangent line is:

y - ln 6 = (1/2)(x - 2) = x/2 - 1
y = x/2 - 1 + ln 6

This crosses the x axis at y = 0

x/2 - 1 + ln 6 = 0
x/2 = 1 - ln 6
x = 2(1 - ln 6)

The answer is 2(1 - ln 6) ≈ -1.5835189

Answer 3

4

Answer 4

Simply take the derivative of the function to be able to find the slipe of the tangent line at a point

dy/dx=1/x

x=2 so the slope is .5

ln(6)=.5x+b

y=.5x+ln6-1

put 0 in for y and get 1-ln6=.5x

then solve

~-1.5835

Answer 5

y = ln(3x)
dy/dx = 3/3x
sub x=2 into 3/3x:
dy/dx=1/2
sub.x=2 into y:
y=ln6
Eqn of tangent: (y-ln6)/(x-2)=1/2
2(y-ln6)=x-2
2y-ln36=x-2
2y=x-2+ln36
x=2y+2-ln36
When the line crosses the x-axis,y=0
x=2-ln36
x=-1.583518938

Answer 6

y' = 1/x = 1/2
x = 2, y = ln 6

y - ln 6 = (1/2)(x-2)

If y = 0, x = 2-2ln 6