# Math Help - Compound increases and logarithms

1. ## 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!

2. Originally Posted by Mellow
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:

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

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

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

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

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

$t = \frac{ln(3)-ln(2)}{ln(1.05)-ln(1.01)}$

3. 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