Question

solve 300e^.80

Answer 1

i dont understand the e part

Answer 2

e is just a number. It is 2.71828.................... It is the base of the natural logaritm ln. There should be a faility on your calculator to raise e to the power of x. Should be cake after that. Let me know if still stuck.

Answer 3

what does the button on the calculator look like

Answer 4

it should just be an \[e ^{x}\]
it is probably a second function of the button for the natural log ln

Answer 5

Find it?

Answer 6

i think

Answer 7

okay so i put in 300c^.80 and i get 240

Answer 8

e*

Answer 9

okay this is the problem I'm trying to do The number of milligrams of a drug that remains in a patient’s system after t hours is given by the function A(t) = Ie^rt. Don was given 300 milligrams of medicine which leaves his bloodstream at a rate of 20%. How much of the medicine remains in his system after 4 hours?

Answer 10

so i plugged it in A(t) = l(300)e^r(.20)(4)
.20(4) = .80
300e^.80
thats where I'm stuck i think I'm doing ti right

Answer 11

so A is the amount of meds in system? Is that \[1e ^{rt}\]?
what is r?

Answer 12

the rate so it would be the 20% so .20

Answer 13

I'm with you now.

Answer 14

use that e^x button we were on about. \[e ^{0.80}\]
then multiply by 300.

Answer 15

and i got 240

Answer 16

I got 668 mg(rounded off)
I'll check again

Answer 17

here is the multiple choice
a-95.87 mg
b- 12.52 mg
c- 134.80 mg
d- 31.63 mg

Answer 18

reading the question again my answer is totally wrong.
i'll have another look

Answer 19

is the function dose x e to the power of r times t?

Answer 20

yeah it would be e^.20(4)

Answer 21

forget it maybe you can help me how to solve this problem
8^(y – 3) = 2^y

Answer 22

\[e ^{0.8}= e ^{4/5} = \sqrt[5]{e ^{4}}\]
We'll look at it this way and see do we have any joy.

Answer 23

th

Answer 24

i just guessed on it its okay

Answer 25

It's bugging me now :-) \[300e ^{0.80}\]= 668
but he couldn't have more meds after 4 hrs than he did at the start.
You might let me know was the function right or was I just making a silly mistake, When you find out like.
Sorry I couldn't get it out.

Answer 26

Yeah i will ill ask my teacher later (: it doesn't make any since to me haha

Answer 27

Cool

Answer 28

do you know how to do this equation 8^(y – 3) = 2^y

Answer 29

What you need to do is get the 8 on the left the same as the 2 on the right

ie 8 = 2^3. then..............

Answer 30

you have\[(2^{3})^{(y-3)}\]
do you have it from there or will I go on?

Answer 31

um i don't even know how you got that o.0

Answer 32

i got the 2^3 so then your take the stuff from the right side also

Answer 33

eight is the same as two cubed so\[(8)^{(y-3)}=(2^{3})^{(y-3)}\]

Answer 34

okay

Answer 35

do you have it from there? I had an answer worked through but I lost my connection and all the work. I'll do it again if you need it.

Answer 36

i don't know how to do it

Answer 37

Stay with me so

Answer 38

i got how you made both sides = now what

Answer 39

so the next step uses exponential laws to give
\[(2^{3})^{(y-3)} = 2^{(3y-9)}\]
can you see that?

Answer 40

where do you get the 9 from

Answer 41

omg i have to go pick up my little sister from school ill be back tho maybe you can put the steps down and when i come back on ill look through them

Answer 42

3(2y-3) = 6y-9 (check out your exponential laws)

Then we have\[2^{(3y-9)} = 2^{y}\]
Are ya with me?

Answer 43

Then you use log laws\[\log _{a}a=1\]
So you get the log to the base 2 of both sides
leaving you with
3y-9=y
so 2y = 9
and y = 2/9

Is that the answer?

Answer 44

the answers are either -4.5, 4.5, -1.5, or 1.5