A radioactive isotope is decaying at a rate of 18% every hour. Currently there are 100 grams of the substance. When will there be only 1 gra

We have

1 = 100e^(-.18t) divide both sides by 100

.01 = e^(-.18t)    take the log of both sides

ln . 01 =ln e^(-.18t)    and since ln e = 1, we can write

ln .01  = (-.18t)       divide both sides by -.18

ln .01 / (-.18)  = t = about 25.58 hrs.

I just wanted to work through this and reproduce the formula for myself - I need the practice.

thanks for teaching me this Chris.  :)

$$\\A=\displaystyle\lim_{t\rightarrow \infty}A_0\left(1-\frac{0.18}{t}\right)^{nt}\\\\
let \;\;\frac{1}{m}=\frac{-0.18}{t} \\\\
so\;\;t=-0.18m
\\A=\displaystyle\lim_{m\rightarrow \infty}A_0\left(1+\frac{1}{m}\right)^{m*-0.18n}\\\\
A=A_0e^{-0.18n}$$