@SolomonZelman

Question

@SolomonZelman

jakob · Answer

attachment

solomonzelman · Answer

Similarly to the previous problem, you are given $N_0$ and $k$.
Can you plug these values and rewrite the function?

jakob · Answer

Can you just guide me step through step I cant even Get to understand this S***

solomonzelman · Answer

You are using 11-gram sample of substance, so you initial substance value is 11 grams. That is, $\color{blue}{N_0=11}$.

solomonzelman · Answer

and the $k{\tiny~}$ value is given explicitly to be $\color{blue}{k=0.1247}$.

solomonzelman · Answer

Can you plug these $\color{blue}{N_0}$ and $\color{blue}{k}$ into the function $\color{blue}{N}$ (that gives the amount of substance at any given time $\color{blue}{t}$)?

jakob · Answer

Yes

solomonzelman · Answer

Ok, what do you get?

solomonzelman · Answer

(What is your function, after you plug in those values?)

jakob · Answer

Hold on sorry

jakob · Answer

N=11e^-01247t 
Right?

solomonzelman · Answer

Ok, good!

solomonzelman · Answer

Well, just in general about the half life (I'll use the general symbols)  ....

solomonzelman · Answer

$\color{blue}{\displaystyle N(t)=N_0e^{kt} }$  is a function that models the substance at
any given time $t$, where $\color{blue}{\displaystyle N_0 }$ is the initial amount of substance.

At what time $t$, would your substance reach its half-life? In other words,
at what time wll the substance be twice smallerthan the initial?
A substance would be twice smaller than the initial amount when
$\color{blue}{N(t)=\frac{1}{2}N_0 }$, in other words you will have to solve $\color{blue}{ \frac{1}{2}N_0=N_0e^{kt} }$
((I obtained this last equation with a substitution))

How do we solve?
$\color{blue}{\displaystyle \frac{1}{2}N_0=N_0e^{kt} }$

$\color{blue}{\displaystyle \frac{1}{2}=e^{kt} }$        (divided by $N_0$ on both sides)

solomonzelman · Answer

So, basically, whenever you want a half life, you just need to find the value of t, for which $\color{blue}{\displaystyle e^{kt}=\frac{1}{2} }$.

Does this make sense? Yes, because when $\color{blue}{\displaystyle e^{kt}=\frac{1}{2} }$, then you multiplying $\color{blue}{\displaystyle N_0 }$ times $\color{blue}{\displaystyle e^{kt} }$, is exactly the same as multiplying times 1/2, and there is the reason why $\color{blue}{\displaystyle e^{kt}=\frac{1}{2} }$ finds the Half-life of a substance.

solomonzelman · Answer

In your case,  $\color{blue}{\displaystyle e^{0.1247t}=\frac{1}{2} }$.

solomonzelman · Answer

Can you solve this equation?

jakob · Answer

.5 would be 1 or do I leave it

solomonzelman · Answer

oh, the exponent is negative.

solomonzelman · Answer

$\color{blue}{\displaystyle e^{-0.1247t}=\frac{1}{2} }$

solomonzelman · Answer

t=0.5? if that is your result, then redo.

jakob · Answer

Whats my answer

solomonzelman · Answer

that is what I should ask you.

solomonzelman · Answer

(not the other way around)

jakob · Answer

Uhhhhh

jakob · Answer

brb

solomonzelman · Answer

$\color{blue}{\displaystyle e^{-0.1247t}=\frac{1}{2} }$

 $\color{blue}{\displaystyle (e^{0.1247t})^{-1}=2^{-1} }$

 $\color{blue}{\displaystyle e^{0.1247t}=2 }$        (a little simplified)

solomonzelman · Answer

$\color{blue}{\displaystyle \ln(e^{0.1247t})=\ln(2) }$ $\tiny \[0.5em]$
 $\color{blue}{\displaystyle 0.1247t\ln(e)=\ln(2) }$$\tiny \[0.5em]$
 $\color{blue}{\displaystyle 0.1247t=\ln(2) }$$\tiny \[0.5em]$
 $\color{blue}{\displaystyle t=(\ln 2)/(0.1247) }$$\tiny \[0.5em]$

that is what would be done with logarithms, and then a calculator (or a tailor series approximation for ln(2) with more willingness),

solomonzelman · Answer

or you can graph,
1.    $\color{blue}{\displaystyle y=e^{0.1247t}}$
2.    $\color{blue}{\displaystyle y=2}$

and find the intersection.

solomonzelman · Answer

(Like I advised to solve the previous problem)

jakob · Answer

K hold on

jakob · Answer

N=11e^-2t

solomonzelman · Answer

I'm not sure I understand why you wrote this.

jakob · Answer

Sorry I just trying to make it easier for me

solomonzelman · Answer

We want to solve
$\color{blue}{\displaystyle N_0/2=N_0e^{-k0.1247t}}$

why? because we want to find the value of t, for which your output substance would be half the initial amount (def. of "half-life").

solomonzelman · Answer

Excuse me, it should be just -kt in the exponent.

solomonzelman · Answer

Then, division by $N_0$ on both sides gives,
$\color{blue}{\displaystyle 1/2=e^{-kt}}$

jakob · Answer

YEAH IDK

solomonzelman · Answer

What is half-life, can you explain in your words?

jakob · Answer

You take time and make it half of it old number