# Determine half-life of this substance

1. Feb 24, 2005

### thomasrules

A substance decays from 20g to 15g in 7h.Determine the half-life of the substance.

I know that: $$M=M_{o}\left\frac{1}{2}\right^\frac{t}{h}$$

Last edited: Feb 24, 2005
2. Feb 24, 2005

### Diane_

I'm insufficiently skilled with Latex (and more than sufficiently lazy not to want to figure it out right now) to post this properly, but there is a small problem with the equation you've posted. The power on the (1/2) should be (t/T), where T = the half-life in units of whatever t is. Given that, you can solve your equation for T, thus:

M = M0(1/2)^(t/T)

ln(M) = ln(M0(1/2)^(t/T))

ln(M) = (t/T)ln(M0/2)

T = t ln(M0/2)/ln(M)

Since you have t in hours, this will give you the half-life in hours.

There are other approaches using more standard exponential decay formulas (decay constants, for instance), but all of them end up with logs eventually.

3. Feb 24, 2005

### thomasrules

ok thank you but by the way whats wrong with my latex:

"$$M=M_{o}\frac{1}{2}^\frac{t}{h}$$"

Last edited: Feb 24, 2005
4. Feb 24, 2005

### vincentchan

you miss the "\" in front of the frac

5. Feb 24, 2005

### thomasrules

Ok I did that and got an answer of 5.95 but thats not the answer

SOMEONE HELP

Last edited: Feb 24, 2005
6. Feb 24, 2005

### vincentchan

this step is wrong!

here is the right one...
$$M=M_{o}\left\frac{1}{2}\right^\frac{t}{h}$$
$$\frac{M}{M_0}=\left\frac{1}{2}\right^\frac{t}{h}$$
$$log_2(\frac{M}{M_0})=log_2(\frac{1}{2}^\frac{t}{h})$$
$$log_2(\frac{M}{M_0})=\frac{t}{h}log_2(\frac{1}{2})$$
$$log_2(\frac{M}{M_0})=\frac{t}{h}log_2(2^{-1})$$
$$log_2(\frac{M}{M_0})=-\frac{t}{h}log_2(2)$$
$$log_2(\frac{M}{M_0})=-\frac{t}{h}$$
$$h=-\frac{t}{log_2(\frac{M}{M_0})}$$