Log-log graphing, slopes, and exponents

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  • Опубліковано 6 вер 2024
  • An introduction to graphing power functions on a log-log graph and a basic example of finding an approximate power-law relationship from a set of data.

КОМЕНТАРІ • 7

  • @nikaroni0
    @nikaroni0 8 років тому +1

    trying to do some data analysis at work with some lab data, and this worked perfect. My data set came out so perfectly linear (R sq. of like 0.9991) and i was so geeked out! I have no idea why this wasn't taught in my high school. I'm glad i know why this works now. Its crystal clear when you used the log rules to convert the exponential equation.

    • @giovannisantostasi9615
      @giovannisantostasi9615 10 місяців тому

      You know Kepler discovered the laws of motion of the planets plotting the distance and time of revolution on a log log graph? Before that people (including Kepler) had all sorts of mystical ideas on how planets moved around the sun (or the earth in the geocentric model). Logs were just invented and out of desperation after trying all sorts of models he decided to use a log log graph and it was astonished when it looked like a perfect straight line.

  • @giovannisantostasi9615
    @giovannisantostasi9615 10 місяців тому

    You know Kepler discovered the laws of motion of the planets plotting the distance and time of revolution on a log log graph? Before that people (including Kepler) had all sorts of mystical ideas on how planets moved around the sun (or the earth in the geocentric model). Logs were just invented and out of desperation after trying all sorts of models he decided to use a log log graph and it was astonished when it looked like a perfect straight line.

  • @Skeletron377
    @Skeletron377 8 років тому

    Great informative, straightforward, easy-to-follow video!

  • @pmgear
    @pmgear 3 роки тому

    Awesome, just what I needed

  • @MA-qz1sd
    @MA-qz1sd 4 роки тому +1

    waited 10 mins for to find A to find you werent doing it

    • @davidmetzler
      @davidmetzler  4 роки тому +1

      Yeah, would have been good to wrap that up. But that's the easy part. Once you have the exponent, just plug in the x,y values for one of the points, and solve for A (just divide).