Find an equation for the tangent line of 𝑦=𝑒^4𝑥 that is parallel to the linear equation 𝑦=5𝑥+15.

Question

Provide your answer in slope–intercept form of a linear equation, 𝑦=𝑚𝑥+𝑏, where 𝑚 is the slope and 𝑏 is the 𝑦- intercept. Express 𝑚 and 𝑏 as exact numbers.

Answer 1

First, find the slope of the line that we are looking for by using derivative:

Given: y = e^4x

To find the slope (m):

m = dy/dx = 4e^4x

The slope of the line that we are looking for and y = 5x+15 have the same slope because they are parallel.

m=5.

5 = 4e^4x

5/4 = e^4x

ln (5/4) = 4x

x = ¼ ln (5/4)

Using the original equation y = e^4x :

y = e^{4•¼ ln (5/4)}

y = e ^{ln (5/4)}

y = 5/4

Using y = mx + b to solve for b:

5/4 = 5 (¼ ln (5/4))+b

b = 5/4 - (5/4) ln (5/4)

Therefore the equation is:

y = 5x + 5/4 - (5/4) ln (5/4)

Answer 2

y = 5x+15 has slope = y' = 5

y=e^4x has slope = y' = 4e^4x

4e^4x = 5

e^4x =5/4 = 1.25

ln1.25 = 4x

x = .25ln1.25

y =e^((4)(.25ln1.25)) = e^ln1.25 = 1.25

slope = 5, through the point (.25ln1.25, 1.25)

1.25 = 5(.25ln1.25) + b

b = 1.25-1.25ln1.25

y = 5x + 1.25 - 1.25ln1.25

5= the difference in y coordinates divided by the difference in x coordinates = (y-1.25)/(x-.25ln1.25)

y-1.25 = 5(x-.25ln1.25)

y = 5x - 1.25ln1.25 + 1.25

or

y = 5x -(5/4)ln(5/4) + 5/4

m=5, b = 5/4-(5/4)ln(5/4)

(5/4)ln(5/4) = approximately 0.279

1.25-0.279 = 0.971

the equation is roughly

y=5x+0.971

Answer 3

y = e^4x

y' = 4e^4x

5 = 4e^4x

0.8 = e^4x

ln 0.8 = 4x

Can you solve for x and answer?