You are starting a new job and are told that your pay will be $3 on the first day. Each day after that, your pay will double until you reach the amount of $96. Starting from the first day, on what day will you reach $96?

[tex]96 = 3\times32 = 3\times2^5 = Ab^t,\\\text{ with }A=3, b=2,\text{ and }t=5[/tex]. A is the starting amount at t=0. b is the base, two because the amount doubles each day. b would be 3 if it tripled every day. t is the final day number, counting from zero.

[tex]A=3,b=2,t=0,Ab^t=3\times2^0=3[/tex]

[tex]A=3,b=2,t=1,Ab^t=3\times2^1=6[/tex]

[tex]A=3,b=2,t=2,Ab^t=3\times2^2=12[/tex]

[tex]A=3,b=2,t=3,Ab^t=3\times2^3=24[/tex]

[tex]A=3,b=2,t=4,Ab^t=3\times2^4=48[/tex]

[tex]A=3,b=2,t=5,Ab^t=3\times2^0=96[/tex]

We can also do this with units included.

Obviously the unit for A is dollars, and t is in days. But what is the unit for b?

Well, the amount doubles once _per day_. So try b = 2 per day = 2/day. That doesn't work. The correct value for b including units is

[tex]b=2^\left(\frac{1}{\text{day}}\right)[/tex].

So if you wanted to express time in weeks while still doubling once per day, you would multiply by a value equivalent to one, [tex]\frac{7\,\text{day}}{\text{week}}[/tex], and cancel units:

[tex]b=2^{\left(\frac{1}{\text{day}}\frac{7\,\text{day}}{\text{week}}\right)}=2^{\left(\frac{7}{\text{week}}\right)}=(2^7)^{\left(\frac{1}{\text{week}}\right)}=128^{\left(\frac{1}{\text{week}}\right)}[/tex].

[tex]A=\$3, b=2^\left(\frac{1}{\text{day}}\right), t=5\,\text{day},\\Ab^t=\$3\big\left(2^\left(\frac{1}{\text{day}}\right)\big\right)^{5\,\text{day}}=\$3\,2^5=\$96[/tex]