pam patterson 3

 

Open thread. Go wild!

(Thanks to Pam for the submission.)

  • Looking for answers — my tweets asking for explanation why ln 2^x was x ln 2 were unanswered; seems the answer to my question would help this student.

  • Logs were one of the most mysterious topics for me while I was in high school-I didn’t understand log rules so I just memorized how to follow them. This student seems to have memorized some themes of “things that happen to symbols” when logs are around. This includes moving numbers in front of a variable to the exponent position and taking logs of both sides. Of course, without meanings for log rules, this student is applying the rules as best they can and the results are haphazard and mathematically incorrect.

    I’ll try to explain SueJ’s questions with a specific example to show why ln(2^3)=3ln(2). I think of ln(2) as “to what power do I have to raise e to to get 2?”

    I think of the answer to ln(2^3) as “to what power do I have to raise e to get 2^3?”

    It turns out that ln(2) is approximately .69 because 2.7^.69 is approximately 2.

    So if i know what power I have to raise e to get 2(namely, .69) I know what to raise e to to get 2^3.

    e^(.69*3) = (e^.69)^3 = 2^3.

    Thus if I’m asking the question, to what power do I have to raise e to to get 2^3 the answer is 3 times as large as what I need to raise e to get 2. Thus the answer is 3 times .69, or equivalently 3 times ln(2).

    Thus ln(2^3)=3ln(2).

    I found this to be absolutely one of the harder things in high school mathematics to understand. It really comes down to seeing a logarithm as an exponent.

  • I think of ln 2^x as asking “e to the what is 2^x” and I turn that into an actual equation, e^t = 2^x.

    Then I want to relate this to ln 2, so I give that a name also, say u, and I write e^u = 2.

    (This is another way of saying the same things Cameron was about thinking of logs as exponents.)

    Now I stare at my two equations, and I think I’m wanting to get both t and u in there somehow, so I notice that I can substitute e^u in place of 2, and get e^t = (e^u)^x and then simplify to e^(ux).

    Then t = ux, or, going back to what those letters stand for, ln(2^x) = ln(2) x.

  • For this particular bit of student work, though, I wonder how to get them to have the reaction to see e^x as a “thing”, maybe calling it a new name with some new variable, so that they recognize this as a quadratic. What kind of pattern recognition is at play there?