Question

Bob's salary grows by $1000 each year. Mary's salary grows by 1% each year. Which one has a salary that grows exponentially?

Answer 1

Mary's increase is exponential, because next year, she will get an increase that is partially based on this year's increase. This pattern will continue in years ahead.

Explanation:

Bob's increase is fixed at $1000, and will not change. This year's increase does not affect next year's increase.

Mary, on the other hand will get 1% more this year. Next year, she will get 1% of 101% (including this year's increase in the new salary). The following year, both increases are taken into account when calculating her increase.

This way in which previous changes contribute to upcoming changes is the characteristic of exponential growth (or decay).

Answer 2

Mary's salary grows exponentially.

Explanation:

Having an exponential salary means the amount it increases by changes each year, and it's in terms of how much money you have at that given year.

This can't be Bob, since it's just $1000 each year. If his salary is $1300 one year, the next it'll be $2300. If it's $235,000 one year, the next it'll be $236,000.

We can model this relationship as a function of time (t), in years, and outputting salary (`S_b`), in dollars:

`S_b=1000t+x`

(`x` is just a placeholder for whatever his starting value is at his first year, which isn't given in this question).

Mary's salary grows by 1% each year. So if one year her salary is $1000, then her salary the next year can be written as:

`$1000*1.01=$1010`

And if her salary is $235,000, the next year, it will be:

`$235,000*1.01=$237,350`

And after two years, it'll be:

`$237,350*1.01`

which can be rewritten as

`$235,000*1.01*1.01`

Or even simplified

`$235,000*(1.01)^2`

Therefore, her salary (`S_m`) after a number of years (`t`) can be written as

`S_m = x*1.01^t`

where `x` is the initial salary. Because the `t` is in the exponent, this makes her salary exponential, while Bob's is linear.