1 / 6

The function e x and its inverse, lnx

y=3 x. y=e x. y=2 x. x. The function e x and its inverse, lnx. The functions like y = 2 x , y = 3 x are called exponential functions because the variable x is the power (exponent or index) of a base number. The graph of y = 2 x , y = 3 x and y = e x.

eric-hull
Download Presentation

The function e x and its inverse, lnx

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. y=3x y=ex y=2x x The function ex and its inverse, lnx The functions like y = 2x, y = 3x are called exponential functions because the variable x is the power (exponent or index) of a base number. The graph of y = 2x, y = 3x and y = ex It is clear from the graph that the number e is somewhere between 2 and 3 but closer to 3 than 2. This is a special number and its value correct to 8 decimal places is e =2.18281828.

  2. y 1 x 1 Graph of ex and lnx y = ex y = x y = lnx y = ex y = lnx

  3. Evaluating function ex and lnx Evaluate (i) e2 (ii) e-3 (iii) ln0.5 (iv) ½ ln10 (i) 7.39 (ii) 0.0498 (iii) 7.39 (iv) 1.15 Find the value of x (i) ln ex = 3 (ii) elnx = 5 (iii) e2lnx = 16 (iv) e-lnx = ½ (i) x = 3 (ii) x = 5 (iii) x = 4 (iv) x = 2

  4. Solving equations involving ex and lnx Solve for x (i) 3e2x – 1 = 54 (ii) 3e2x –5ex = 2  3e2x = 55 Let y = ex  e2x = 18.333..  3y2 – 5y = 2  2x= ln18.333..  3y2 – 5y – 2 = 0  2x= 2.9087..  (3y )(y )  x= 1.45  (3y + 1)(y - 2)  y = - 1/3 or y = 2  ex = 2  x = 0.693

  5. Solving equations involving ex and lnx Solve for x (i) ln(3x – 5) = 3.4 (ii) ln(3x + 1) – ln3 = 1  ln((3x + 1)/3) = 1  3x – 5 = e3.4  ((3x + 1)/3) = e1  3x – 5 = 29.964…  ((3x + 1)/3) = 2.718…  3x = 34.964…  x= 11.7.  3x + 1 = 8.1548..  3x = 7.1548… x = 2.38

  6. Exponential Decay/Growth A quantity N is decreasing such that at time t (a) Find the value of N when t = 5 (b) Find the value of t when t = 2 = 18.4 (a) (b) t = 16.1

More Related