what is the soloution to this quetion?

Question

Substance-151 has a half-life of 3.2 days.
A sample originally has a mass of 740 mg. Find the formula for the mass remaining after t days.

if you can please exlpain step by step

Answer 1

Where t is in days, the amount A(t) remaining at time t from the start (at t = 0) is given by

A(t) = A(0) e^( - (ln 2) t / 3.2) mg = 740 e^( - (ln 2) t / 3.2) mg ......(1)

[Therefore, after 1 day, for example, A(1) = 740 e^( - 0.693147181... (1) / 3.2) mg. = 595.8814... mg; after 3.2 days, A(3.2) = 740 e^(- 0.693147181... 3.2 / 3.2) = 740 / 2 = 370 mg,, as advertised.]

O.K., now let's justify equation (1).

We start by simply writing down a form for exponential decay, with an "e-folding decay-timescale" t0 (the time for the sample to be reduced by a factor of ' e '):

N(t) = N(0) e^(- t / t0). ......(2)

Then, when N(t) = 1/2 N(0) [which defines the "half-life" t_half],

(1/2) N(0) = N(0) e^ (- t_half / t0), so that e^ (+ t_half / t0) = 2, and thus :

t_half = t0 ln (2), or t0 = t_half / ln (2). ......(3)

Therefore, in eqn. (2),

N(t) = N(0) e^(- ln (2) t / t_half) = 740 e^(- ln (2) t / 3.2) mg.

This justifies the formula with which I began.

Live long and prosper.

Answer 2

use y=a*b^x
where a is what you start with -----740
b is what happens to it ----- 1/2 or .5
and x is how often t/3.2
so y=740*0.5^(t/3.2)

and I tried to figure out your other problem about 14hour and 1000 and 20 hour and 70000...with no luck

I started with 1000=ab^14
and 70000=ab^20 but couldn't think of how to figure it ...it's also almost 1am! so brain not working too well
sorry