Decaying logarithms help needed.?

Question

chemical has initial mass of 1000g, reduced by 10% per second.
-first need a formula, giving mass (m) and time (t) secs, and rearrange to make t subject. What remains after 10 secs? How long to the nearest sec until mass is 10g. After how many secs is the mass less than 1 gram?
Step by step help will be greatly appreciated, it would really help me out with understanding the principles.

Answer 1

The decay formula is
M = A*e^(-kt) where A is the initial mass, t is the elapsed time, and k is some constant.

Clearly A = 1000, and since it's reduced to 9000 g in 1 second you substitute M = 900, t = 1 in the equation to get

900 = 1000*e^(-k),
which I hope you can solve to find k.


[It's k = - ln(9/10)]

Using this value of k in the formula, substitute t = 10 to answer the first question. For the second, put M = 10 and solve for t.

Finally solve 1000*e^(-kt) < 1,
but after you've divided both sides by 1000 and then taken logs, don't forget that when you then divide by -k you reverse the < to get t > ... whatever.

Answer 2

You don't say what level of math you're at, and the calculus in a couple of those answers might be too much for you.

If the chemical loses 10% of mass every second, 90% is left, so a simple Alg 2 equation would be m = 1000(0.9)^t. You'll find that e^(-k) in the sophisticated version turns out to be 0.9.

So after 10 seconds m = 1000(0.9)^10 = 348.7

To solve 10 = 1000(0.9)^t for t, take the log (common or natural, doesn't matter, but the 10 and the 1000 work well with common) of both sides after dividing by 1000:

0.01 = 0.9^t
log .01 = log ( 0.9^t)
-2 = t log 0.9
t = -2 / log 0.9 = -2 / -0.0458 = 43.7 s.

for m<1g,
0.001 = 0.9^t
-3 = t log 0.9
t = -3 / log 0.9 = -3 / -0.0458 = 65.6 s.

Answer 3

m0 = 1000; initial mass
(1/m)(dm/dt) = -0.1; 10% per sec decay

>> d(Ln (m))/dt = -0.1
>> Ln(m) = Ln(m0) - 0.1 t; Since at t = 0, m = m0.
>> m = exp(-0.1t + Ln(m0))

What remains after 10 secs? substitute t=10, m0 = 1000; this gives m10 ~ 367.

How long does it take to make the mass < 1gm?
put m = 1, >> for this time t,
-0.1t + ln(m0) = 0
t = 69.07
So, at t = 70 secs, it will be less than 1 gm

Answer 4

fraction (1/e) remains after 10 seconds
(e = 2.71828182...)

amt remaining = 1000 * e^-(t/10)

e^-(t/10) = (1/1000)
e^(t/10) = 1000
(t/10) = ln (1000) to base e
t = 10 * (ln (1000) to base e)

t = 30 * ln(10) to base e
so 60 < t < 90 because 2 < (ln(10) to base e) < 3
...........................
theory of it
(1 +(1/n))^n = e (limiting value for large n)