whats the exponential decay equation for 100 atoms that decay 1/20 per day

1/20 = .05

dn/dt = -.05 n

dn/n = -.05 dt

ln n = -.05 t + c

n = C e^-.05 t

n = 100 e^-.05 t

Hmm. Since the amount present on day n is 0.95 times the amount on day (n-1) I think the equation should be

n = 100*0.95t
= 100 * eln.95 t

ln.95 = -.05129

I keep thinking this is absorbed into the c of the integral, but that must certainly be 100.

How to reconcile?

To find the exponential decay equation for a given situation, we need to use the formula:

N(t) = N₀ * e^(-kt)

Where:
- N(t) represents the number of atoms remaining at time t
- N₀ represents the initial number of atoms
- e is the base of natural logarithm (approximately equal to 2.71828)
- k is the decay constant

In this case, you have 100 atoms that decay at a rate of 1/20 per day. To find the decay constant (k), we need to determine the value where N(t) decreases by 1/20 (0.05) per day.

So, we can set up the equation:

N(t) = N₀ * e^(-kt)

N(t + 1 day) = N₀ * e^(-k(t + 1))

From the given information, we know that N(t + 1 day) = N₀ - 1/20 * N₀ = (20 - 1)/20 * N₀ = 19/20 * N₀

Now, let's substitute these values into the equation:

N₀ * e^(-k(t + 1)) = 19/20 * N₀

Canceling out N₀ on both sides:

e^(-k(t + 1)) = 19/20

Taking the natural logarithm (ln) of both sides:

ln(e^(-k(t + 1))) = ln(19/20)

Simplifying:

-k(t + 1) = ln(19/20)

Solving for k:

k = -ln(19/20)/(t + 1)

Given that t = 0 (initial time), we can further simplify:

k = -ln(19/20)/1
k = ln(20/19)

Therefore, the exponential decay equation for this scenario is:

N(t) = 100 * e^(-ln(20/19) * t)