Results 1 to 5 of 5

Math Help - Logs

  1. #1
    Newbie
    Joined
    Aug 2011
    Posts
    16

    Logs

    Need some help to tell me what I'm doing wrong.

    solve for y:

    2e^(-y/3) = t + 4
    ln(2e^(-y/3)) = ln(t+4)
    -y/3(2) = ln(t+4)
    y= -(3/2)ln(t+4)

    ln(20-y) = -.2t + ln8
    *Mult both sides by e
    20-y = 8e^(-.2)
    y= 20 - 8e^-.2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2

    Re: Logs

    One thing at a time...

    2\cdot e^{-y/3} = t+4

    Try simple things first. Divide by 2

    e^{-y/3} = (t+4)/2

    Now the logarithm.

    -y/3 = ln\left(\frac{t+4}{2}\right)

    Can you finish?

    On the second, think really, REALLY hard about "multiply by e" to remove a logarithm. Until you see that this is very, VERY wrong, keep thinking about it.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2011
    Posts
    16

    Re: Logs

    I'll continue on the 2nd but for the first I just got y = -3ln((t+4)/2) Which I still think is wrong.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,110
    Thanks
    2

    Re: Logs

    No, that's good, but it can be written a few other ways. Figure out how to get this one:

    y = 3\cdot ln\left(\frac{2}{t+4}\right)

    You should also worry about the Domain.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466

    Re: Logs

    Quote Originally Posted by crazycarlos61 View Post
    Need some help to tell me what I'm doing wrong.

    solve for y:

    2e^(-y/3) = t + 4
    ln(2e^(-y/3)) = ln(t+4)
    Here is your error: ln(ab)= ln(a)+ ln(b), NOT a ln(b). ln (2e^{-y/3})= ln(2)+ ln(e^{-y/3})= ln(2)- y/3, not "-y/3(2)". To solve for y, you would then subtract ln(2) from both sides which would give you a factor of 2 in the denominator inside the logarithm. That is, this is the same as first dividing both side by 2 as TKHunny suggested.

    -y/3(2) = ln(t+4)
    y= -(3/2)ln(t+4)

    ln(20-y) = -.2t + ln8
    *Mult both sides by e
    NO! Not "multiply both sides by e". Take "e to the power of both sides" or "take the exponential of both sides."
    What you did is correct but thinking "multiply" when you don't mean that will lead you into errors eventually.

    20-y = 8e^(-.2)
    y= 20 - 8e^-.2
    What happened to the "t"? exp(-.2t+ ln(8))= exp(-.2t)(8)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: February 22nd 2011, 05:39 PM
  2. Logs
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 24th 2010, 07:52 AM
  3. Logs
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 10th 2009, 06:08 PM
  4. Dealing with Logs and Natural Logs
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 14th 2008, 06:18 AM
  5. several questions-logs/natural logs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 12th 2007, 08:58 PM

Search Tags


/mathhelpforum @mathhelpforum