Results 1 to 3 of 3

Math Help - Compound increases and logarithms

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    2

    Compound increases and logarithms

    Hello!

    I'm stuck on what I think should be a relatively simple question. A person has a choice of two jobs. Job A starts at 20000 pa, but increases at a rate of 5% a year. Job B starts at 30000 pa, but increases at 1% a year.

    I would like to know after how many years (i.e. for what value of t) the salary of Job A will overtake that of Job B. I've tried to turn it into an inequality as follows, where t is the number of years:

    20000 x 1.05^t > 30000 x 1.01^t

    My guess is that something needs to be done with logarithms in order to bring t down from being a power. But no matter what I do, I seem to come up with the rather infuriating and obvious conclusion that "202.98t > 43.09t"!

    Any help or suggestions as to how I can find the value of t which fulfills the inequality?

    Thank you so much for your help in advance. I've been pulling my hair out for a while, and have a feeling I will kick myself when I find out the answer! >_< I've found the answer by trial and error, but would like an exact value for t and some proper mathematical technique if possible.

    Mellow

    EDIT: Sorry if this is in the wrong forum. I chose the one I think my dilemma fits into best. Mods, please feel free to shift it around!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by Mellow View Post
    Hello!

    I'm stuck on what I think should be a relatively simple question. A person has a choice of two jobs. Job A starts at 20000 pa, but increases at a rate of 5% a year. Job B starts at 30000 pa, but increases at 1% a year.

    I would like to know after how many years (i.e. for what value of t) the salary of Job A will overtake that of Job B. I've tried to turn it into an inequality as follows, where t is the number of years:

    20000 x 1.05^t > 30000 x 1.01^t

    My guess is that something needs to be done with logarithms in order to bring t down from being a power. But no matter what I do, I seem to come up with the rather infuriating and obvious conclusion that "202.98t > 43.09t"!

    Any help or suggestions as to how I can find the value of t which fulfills the inequality?

    Thank you so much for your help in advance. I've been pulling my hair out for a while, and have a feeling I will kick myself when I find out the answer! >_< I've found the answer by trial and error, but would like an exact value for t and some proper mathematical technique if possible.

    Mellow

    EDIT: Sorry if this is in the wrong forum. I chose the one I think my dilemma fits into best. Mods, please feel free to shift it around!
    That inequality looks good to me. You can cancel a factor of 10,000 immediately and I'd make it an equation to find when the two are equal:

    <br />
2 \times 1.05^t = 3 \times 1.01^t

    You can then take the logs of both sides and recall the laws of logs: log_c(ab) = log_c(a) + log_c(b). You should then be able to bring down the t.

    Spoiler:
    ln(2 \times 1.05^t) = ln(2) + ln(1.05^t) = ln(2) + t\, ln(1.05)

    ln(3 \times 1.01^t) = ln(3) + ln(1.01^t) = ln(3) + t\, ln(1.01)

    t\, ln(1.05) + ln(2) = ln(3) + t\, ln(1.01)

    t\, ln(1.05) - t\, ln(1.01) = ln(3) - ln(2)

    t\,(ln(1.05)-ln(1.01)) = ln(3)-ln(2)

    t = \frac{ln(3)-ln(2)}{ln(1.05)-ln(1.01)}
    Last edited by e^(i*pi); August 23rd 2009 at 05:32 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    2
    Thank you so much e^(i*pi)! I had completely forgotten about the laws of logs, but it seems a lot clearer now. I'll take some time to go back over the laws and remind myself of the rest of them now :P
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. percentage increases
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: September 17th 2011, 10:38 AM
  2. Entropy increases as number of bins increases
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: February 24th 2011, 01:14 PM
  3. Percentage increases
    Posted in the Algebra Forum
    Replies: 5
    Last Post: May 26th 2008, 12:17 PM
  4. graph that increases or decreases
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 9th 2006, 01:33 AM
  5. Calculating average annual increases
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: June 29th 2006, 07:19 PM

Search Tags


/mathhelpforum @mathhelpforum