1. Originally Posted by BlueStar
...
e^3x = 6 (divide the other 3?)
...

the 3 on the left is a power, how did you manage to divide by it?!

2. What do I do then? That's what I'm asking.

3. Originally Posted by BlueStar
What do I do then? That's what I'm asking.
take the log of both sides and bring the power down.

$\displaystyle e^{3x} = 6$

$\displaystyle \Rightarrow \ln e^{3x} = \ln 6$

$\displaystyle \Rightarrow 3x \ln e = \ln 6$

$\displaystyle \Rightarrow 3x = \ln 6$

$\displaystyle \Rightarrow x = \frac { \ln 6}{3}$

This is my 27th post!!!!

4. Thanks!! That's been bugging me for a while.

5. Now for application:

Radioactive iodine is used to determine the health of the thyroid gland. It decays according to the equation y = ae^-0.0856t, where t is in days. Find the approximate half-life of this substance.

So y = 1e^0.0856t (1 is the whole amount of the substance)

Am I supposed to plug in an arbitrary number of days for t and find y, or make 1 into 0, or what?

6. Originally Posted by BlueStar
Now for application:

Radioactive iodine is used to determine the health of the thyroid gland. It decays according to the equation y = ae^-0.0856t, where t is in days. Find the approximate half-life of this substance.

So y = 1e^0.0856t (1 is the whole amount of the substance)

Am I supposed to plug in an arbitrary number of days for t and find y, or make 1 into 0, or what?
the long, hard systematic way to do this is assume you start with 1 unit, then after the half-life you would end up with 1/2 units, so you simply solve:

$\displaystyle \frac {1}{2} = e^{-0.0856t}$ for $\displaystyle t$

the easier way is to know that for the decay equation in the form $\displaystyle y = A_0 e^{-rt}$ the half-life is given by:

$\displaystyle t_h = \frac { \ln 2}{r}$

7. Originally Posted by BlueStar
Now for application:

Radioactive iodine is used to determine the health of the thyroid gland. It decays according to the equation y = ae^-0.0856t, where t is in days. Find the approximate half-life of this substance.

So y = 1e^0.0856t (1 is the whole amount of the substance)

Am I supposed to plug in an arbitrary number of days for t and find y, or make 1 into 0, or what?
First, for a new question create a new thread.

Second, please get in the habit of using parenthesis for these. The function you want is
y = 1e^(-0.0856t)
(You left out the minus sign as well!)

Okay, now to your question. You want the "half-life" so you wish to know for what t value is y = (1/2)a.

$\displaystyle \frac{1}{2}a = ae^{-0.0856t}$

$\displaystyle \frac{1}{2} = e^{-0.0856t}$

$\displaystyle ln \left ( \frac{1}{2} \right ) = -0.0856t$

Now, $\displaystyle ln \left ( \frac{1}{2} \right ) = ln(2^{-1}) = -ln(2)$ so...

$\displaystyle -ln(2) = -0.0856t$

$\displaystyle t = \frac{ln(2)}{0.0856} \approx 8.09751$

where t is in whatever time unit you are using, in this case days.

-Dan

PS I think the "long hard way" of doing this problem is probably the most beneficial right now. But then, I always do.

8. Originally Posted by topsquark
PS I think the "long hard way" of doing this problem is probably the most beneficial right now. But then, I always do.
i agree, especially for students who have little experience with the material. the manipulation is important for them to learn.

there are a lot of things in math i usually do the hard way as opposed to memorizing formulas (for some reason i find it easier to learn how to work something out than memorize a formula), but this isn't one of them

9. Originally Posted by topsquark
First, for a new question create a new thread.
Sorry.

Okay, now to your question. You want the "half-life" so you wish to know for what t value is y = (1/2)a.

$\displaystyle \frac{1}{2}a = ae^{-0.0856t}$

$\displaystyle \frac{1}{2} = e^{-0.0856t}$

$\displaystyle ln \left ( \frac{1}{2} \right ) = -0.0856t$

Now, $\displaystyle ln \left ( \frac{1}{2} \right ) = ln(2^{-1}) = -ln(2)$ so...

$\displaystyle -ln(2) = -0.0856t$

$\displaystyle t = \frac{ln(2)}{0.0856} \approx 8.09751$

where t is in whatever time unit you are using, in this case days.

-Dan

PS I think the "long hard way" of doing this problem is probably the most beneficial right now. But then, I always do.
Doesn't seem any harder or longer than any other equations. In fact I get that one better. Thanks guys.

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