# Thread: half life word problem using logs

1. ## half life word problem using logs

a 50-mg sample of cobalt-60 decays to 40mg after 1.6min. determine the half-life of the cobalt-60.

this is as far as i got before getting stuck

M = C $(\frac{1}{2})^\frac{t}{h}$

40 = 50 $(0.5)^\frac{1.6}{h}$

0.8 = $(0.5)^\frac{1.6}{h}$

$\frac{1.6}{h}$ = $\frac{log0.8}{log0.5}$

then right here i get stuck and can't get the correct answer. can somone tell me what i'm doing wrong?

i'm getting the answer 0.52 mins, when its susposed to be 5 mins

2. Originally Posted by extraordinarymachine
a 50-mg sample of cobalt-60 decays to 40mg after 1.6min. determine the half-life of the cobalt-60.

this is as far as i got before getting stuck

M = C $(\frac{1}{2})^\frac{t}{h}$

40 = 50 $(0.5)^\frac{1.6}{h}$

0.8 = $(0.5)^\frac{1.6}{h}$

$\frac{1.6}{h}$ = $\frac{log0.8}{log0.5}$

then right here i get stuck and can't get the correct answer. can somone tell me what i'm doing wrong?

Edit: your working is actually fine - you just need to use some arithmetic to make h the subject. I would take the reciprocal and multiply by 1.6

$\frac{h}{1.6} = \frac{log(0.5)}{log(0.8)} \: \: \rightarrow \: \: h = 1.6 \, \frac{log(0.5)}{log(0.8)} = 5.0$ (2sf)

Use the exponential decay formula

$A=A_0e^{-kt}$

$40=50e^{-1.6t}$

$-1.6k = ln(0.8)$

$k = -\frac{1}{1.6}\, ln(0.8)$

$t_{1/2}=\frac{ln(2)}{k} = -\frac{1.6\, ln(2)}{ln(0.8)} = 5.0$ (2sf)

(see spoiler for why this is so)

Spoiler:
Take the exponential decay equation: $A=A_0e^{-kt}$

Half life ( $t_{1/2}$) is defined as the time it takes for half the original substance to decay (ie: $A=0.5A_0$)

$\frac{0.5A_0}{A_0} = e^{-kt_{1/2}}$

A_0 cancels and then take the log. As $0.5=2^{-1}$ use the log power law to say that $ln(0.5) = -ln(2)$

$-ln(2)=-kt_{1/2}$

Divide by $-k$

$t_{1/2}=\frac{ln2}{k}$