basic exponential equation question. please help?

Plugging in the given numbers, N(1) = a(b^1) = 80 => a*b = 80.

and 100mins = ~1.67 hrs (really 1&2/3 hrs) , so N(1.67) = a(b^1.67) = 320.

If I solve the first equation for a, I get a = 80/b. Substituting that into equation 2 gives me

(80/b)(b^1.67) = 320.

move the 80/b to the other side so that the b^1.67 is by itself. => b^1.67 = 320(b/80) =>b^1.67 = 4b

--------This is where it gets tricky-----------

(When in doubt take the log of both sides)

Here we're going to need logarithms, and you will want to use log base b to cancel as much as you can. to denote log base b I will write log{b}.

So, log{b}(b^1.67) = log{b}(4b) => 1.67*log{b}(b) = log{b}(4) + log{b}(b) => 1.67 = log{b}4 + 1

=> log{b}(4) = 2/3 (Note that the real decimal is not 0.67 it is a repeating decimal that can be rewritten as 2/3)

=>log{b}(4) = 2/3 => b^(2/3) = 4 => b = 4^(3/2) => b = 8

Now that we've solved for b, we can plug that into the first equation (a*b=80) to get a.

a(8) = 80 => a=10.

This we can plug into the given formula N(t) = 10(8^t).

---------DONE---------

To check, plug your given information back into your solution.

N(1) = 10(8^1) = 80 (good)

N(1.67) = 10(8^1.67) = 320 (good)

Starting time = 1 hour = 60 mins

Starting value = 80

Next time = 1 hour 40 mins = 100 mins

next value = 320

Growth = 320-80/100-60 = 6 per minute.

So growth rate per hour = 6*60 = 360 per hour.

N(t) = 80(360^t)

n=ab^t

80=a*b^60 and 320=a*b^100

ln80=lna+60lnb and ln320=lna+100lnb

solve this system for lna and lnb

subtract eq1 from eq2: ln320-ln80= (100-60)lnb or ln4=40lnb and lnb=(ln4)/40

5*eq1: 5ln80=5lna + 300lnb

3*eq2: 3ln320= 3lna + 300lnb

subtract them: 5ln80-3ln320= 2lna; ln[80^5/320^3]=2lna; lna=(1/2)ln(80^5/320^3)

Now since lna=(1/2)ln(80^5/320^3), a=sqrt(80^5/320^3) or 10

And lnb=(ln4)/40 and b=4^(1/40) or 1.035265

So we have n= 10*1.035265^t

This answer can be verified by substituting in values of 60 and 100 for t. It checks out.