Radioactive iodine is used to determine the health of the thyroid gland. It decays according to the equation where t is in days. Find the half-life of this substance. Round to the tenths place.For the half life: The half life is the solution (t) of the equation : a/2= ae ^ (-0.0856t)

Question

Radioactive iodine is used to determine the health of the thyroid gland. It decays according to the equation  where t is in days. Find the half-life of this substance. Round to the tenths place.For the half life: The half life is the solution (t) of the equation : a/2= ae ^ (-0.0856t)

jazzyfa30 · Answer

@nincompoop

anonymous · Answer

typ there sorry

anonymous · Answer

solve 
$$\large e^{-.0865t}=.5$$

jazzyfa30 · Answer

?????? idk what that is

anonymous · Answer

that's better
takes only two steps 
$$-.0865t=\ln(.5)$$
$$t=\frac{\ln(.5)}{-.0865}$$

jazzyfa30 · Answer

ok can you walk me through the set up

anonymous · Answer

you have a typo in the problem i think

Radioactive iodine is used to determine the health of the thyroid gland. It decays according to the equation MISSING EQUATION HERE where t is in days.

anonymous · Answer

i assumed the equation was 
$$\large A_0e^{-.0865t}$$

anonymous · Answer

is that correct?

jazzyfa30 · Answer

i copied and pasted the entire Q

anonymous · Answer

yeah, but take a look
i think when you copied and pasted it missed a piece

jazzyfa30 · Answer

thats the way they had it on the Q see

anonymous · Answer

it doesn't matter, i can walk you through this, it is not too hard, and takes really only two steps to solve

jazzyfa30 · Answer

ok

anonymous · Answer

at the very bottom you have this equation
$$\large a/2=ae^{-.0865t}$$ right?

jazzyfa30 · Answer

yea

anonymous · Answer

the $a$ is unnecessary, because the first step in solving is to divide both sides by $a$ and get 
$$\large e^{-.0865t}=\frac{1}{2}=.5$$

anonymous · Answer

so far so good?  we just cancelled the $a$ from both sides of the equation

jazzyfa30 · Answer

yea im following

anonymous · Answer

ok our goal to to get $t$ by itself
at the moment it is in the exponent
the way to get it out of the exponent is to "take the natural log" of both sides, 
do you know what i am talking about?  "no" is a fine answer, i am just wondering

jazzyfa30 · Answer

im glad u stated that because no

anonymous · Answer

ok no problem

anonymous · Answer

the way to solve $e^x=y$ for $x$ is by writing in equivalent logarithmic form as 
$$x=\ln(y)$$
you have two logs on your calculator, $\log(x)$ which is base 10 and $\ln(x)$ which is base $e$
since $e$ is our base, we clearly want to use $\ln(x)$

anonymous · Answer

in any case we will solve using wolfram, not a calculator
so starting with 
$$\large e^{-.0865t}=.5$$ we rewrite as 
$$-.0865t=\ln(.5)$$

anonymous · Answer

now the only think you need to do to get $t$ by itself is to divide both sides by $-.0865$
final answer is 
$$\large t=-\frac{\ln(.5)}{.0865}$$

anonymous · Answer

looks like about 8 http://www.wolframalpha.com/input/?i=-ln%28.5%29%2F.0865

anonymous · Answer

not sure how clear that was, but i hope you see it was really only two steps

jazzyfa30 · Answer

no thats incorrect

anonymous · Answer

it does say "round to the tenths place", right?

anonymous · Answer

ah damn, i put it $.0865$ instead of $.0856$

anonymous · Answer

http://www.wolframalpha.com/input/?i=-ln%28.5%29%2F.0856 try $8.1$

jazzyfa30 · Answer

nope