# Determine half-life of this substance

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

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

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Diane_
Homework Helper
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.

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

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

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you miss the "\" in front of the frac

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

SOMEONE HELP

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ln(M) = ln(M0(1/2)^(t/T))
ln(M) = (t/T)ln(M0/2)
this step is wrong!

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