1 / 11

Using logs to Linearise the Data

Using logs to Linearise the Data. The equation is y = ab x. This graph is NOT linear . Using logs to Linearise the Data. The equation is y = ab x. log y = log(ab x ) log y = log a + logb x Using the addition rule log(AB) = logA + logB

myra
Download Presentation

Using logs to Linearise the Data

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. Using logs to Linearise the Data The equation is y = abx This graph is NOT linear

  2. Using logs to Linearise the Data The equation is y = abx log y = log(abx) log y = log a + logbxUsing the addition rule log(AB) = logA + logB log y = log a + (xlogb)Using the drop down infront rule log y = (logb) x + logaRearranging to match with y = mx + c Take logs of both sides

  3. log y = (logb)x + logaRearranging to match with y=mx + c Matching up : Y axis = log y gradient =m = logb x axis = x C = log a So make a new table of values x = x Y = logy

  4. Plot x values on the x axis and logy values on the y axis

  5. The equation of the line is y = -0.0522x + 2.0973 Y m X c log y = logb x + loga Matching up : y = mx + c gradient = log b = -0.0522 C = log a = 2.0973

  6. gradient = log b = -0.0522 To find b do forwards and back b  log it = –0.0522 Backwards –0.0522  10 it  b b = 10–0.0522 = 0.8867

  7. y intercept = log a = 2.0973 To find a do forwards and back a  log it = 2.0973 Backwards 2.0973  10 it  a a = 102.0973 = 125.1

  8. The exponential equation is y = abx y = 125.1×0.887x

  9. Using the equation y = 125.1×0.887x If x = 5.5 find y y = 125.1×0.8875.5 = 64.7 Check if the answer is consistent with the table x = 5.5 find y y = 64.7 which is consistent with the table

  10. Using the equation y = 125.1×0.887x If y = 65 find x 65 = 125.1×0.887x Take logs of both sides log65 = log(125.1×0.887x) = log(125.1)+log(0.887x) = log(125.1)+xlog(0.887) Using the addition rulelog(AB) = logA + logB Using the drop down infront rule

  11. log 65 = log(125.1)+xlog(0.887) Forwards and backwards x ×log0.887  +log 125.1 = log65 log65 –log 125.1  ÷log0.887 = x x = 5.46 Check if the answer is consistent with the table y = 65 find x x = 5.46 which is consistent with the table

More Related