1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Logarithmic and exponential equations

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data
    (2^x - 2^-x)/3=4


    2. Relevant equations

    Using log or exponential rules

    3. The attempt at a solution

    First multiply both sides by 3 so 2^x-2^-x=12
    I thought I could take the log of both sides then condense the log, but that is not right.

    I also attempted to rewrite as 2^x - 1/(2^x) = 12 but I could nowhere with that.

    I also attempted to do (2 - 2^-1)^x = 12
    (2-1/2)^x = 12
    (3/2)^x = 12

    log(3/2)^x = log 12

    x(log 3/2) = log 12

    x= log 12/log 3/2 = 6.128

    But that is not the right answer it is supposed to be 3.595 . Where did I go wrong?
     
  2. jcsd
  3. Dec 9, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can't do most of those things. 2^x - 1/(2^x) = 12 is a good start. Put u=2^x. Then 2^(-x)=1/u. Form an equation for u, solve for it and then find x.
     
    Last edited: Dec 9, 2013
  4. Dec 9, 2013 #3
    Ok, so u - 1/u = 12

    so u^2 - 12u - 1 = 0

    Do I use the quadratic equation to solve for u?

    Sorry, but this is unlike any of the other equations we did.
     
  5. Dec 9, 2013 #4

    Mentallic

    User Avatar
    Homework Helper

    Well, if you were given that quadratic to solve, what would you do?

    Then it's probably one of the harder questions that tests your ability to apply other methods you've previously learnt to solve the problem.
     
  6. Dec 9, 2013 #5
    OK, at first I did not think I should do that, because I would wind up with two answers, but one is negative, which obviously is an extraneous answer because logs cannot be negative. So the other answer was 12.08 so the log 12.08/log 2 = 3.585 which is close to the book's answer.
     
  7. Dec 9, 2013 #6

    Mentallic

    User Avatar
    Homework Helper

    Precisely :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Logarithmic and exponential equations
  1. Exponential logarithm (Replies: 12)

Loading...