radium decomposes at a rate proportional to the amount instantaneously present.Find the half-life of radium if three-fourths of it is present after 8 hours

Answer 1

If y = f(x) then you have
dy/dx = ky
dy/y = k dx
ln y = kx + ln c
y = c e^(kx)
c is the initial amount present (e^0 = 1)
Now, e^(ln 1/2) = 1/2, so we can fold that into the k value, giving
y = c (1/2)^(kx)
since we want to find the half-life.
y(8)/y(0) = (1/2)^(8k) = 3/4
8k = ln(3/4)/ln(1/2)
k = 0.05188 = 1/19.27
so y = c (1/2)^(x/19.27)
and the half-life is 19.27 hours

Answer 2

where does the y(8)/y(0) also 8k comes from?

Answer 3

huh? That is the fraction left after 8 hours!
y(0) is the original amount
y(8) is the amount at time x=8

Answer 4

i got it awhile ago, thank u for answering!

Answer 5

y(8)/y(0) = (1/2)^(8k) = 3/4
especially this part y(8)/y(0) does it need to be in fraction? is it just y only since it is y= ce^kx
i kinda don't get this part but i know that there's a given 3/4 can u please elaborate it i just wanna know and learn tnx.