Question

A 15-g sample of radioactive iodine decays in such a way that the mass remaining after t days is given by m(t) = 15e^-0.087t, where m(t) is measured in grams. After how many days is there inly 5 g remaining?

I set 5 = to the equation and it didnt work:(

Answer 1

thats good start... :)

5 = 15e^(-.087t)
1/3 = e^(-.087t)

take ln both sides

ln(1/3) = -.087t
-1.1 = -.087t
t= ?

Answer 2

So I guess I just forgot to remove 15 by dividing it by 5 again……..stupid I should have looked before I posted this one...

Answer 3

its okay hehehe :)

Answer 4

I have another question?!?!

Answer 5

post

Answer 6

A man invested $5,000 into an account that pays 8.5% per year compounded quarterly.
Find how long it will take for the investment to double….




I just get confused on where to put certain things and exactly what to use.

Answer 7

Final amount = P ( 1 + r/n)^(n*t)

Answer 8

you familiar with above formula

Answer 9

?

Answer 10

Sorry !! I was working on another problem...

Answer 11

Yes I have seen that before.

Answer 12

np

\(\huge \textbf{Final amount} = P(1+\frac{r}{n})^{r*t} \)

P : initial amount = $5000
r : interest rate per year = .085
n : num. of compounding times per year = 4
t : time in years = ?????????????????????????????????????
Final amount : 10,000

Answer 13

can you find t now

Answer 14

Okay! thank you so very much!!!

Answer 15

yw :)

Answer 16

:)

Answer 17

Still there I need help with these equations, i must be missing some rule….

Answer 18

sure ask

Answer 19

a) 5^x = 4^x+1



b) 2^2x+1 = 3^x-2



c) 50/1+ e^-x = 4



d) log x + log(x-1) = (4x)

Answer 20

Ok. you wanto understand how to solve x is it

Answer 21

Yes

Answer 22

23 and 25 says to find the solution...

Answer 23

hmm who is 23 and 25

Answer 24

For #23 and #25 Directions say to find the solution..

Answer 25

those four equation you can solve using only two log properties

Answer 26

which ones?

Answer 27

\(\huge 1) \ \log a^m = m\log a\)
\(\huge 1) \ \log a + \log b = \log ab\)

Answer 28

hmm I know those two but I have tried these problems a couple times and I just cant do them!!!

Answer 29

its okay takes time to understand them.

lets try and solve this
a) 5^x = 4^x+1

Answer 30

5 ln x = x ln x+1

Answer 31

thennn…… divide by 5ln / 4 ln

Answer 32

wait. let me do the first one showing u step by step.
rest three of them u can try ok

Answer 33

kkkk

Answer 34

\(5^x = 4^{x+1}\)

taking log both sides

\(\log(5^x) = \log(4^{x+1})\)
\(x \log(5) = (x+1)\log(4)\)
\(x \log(5) = x\log(4) + \log(4)\)
\(x(\log5 - \log4) = \log(4)\)
\(x = \frac{\log 4}{\log 5 - \log 4}\)

Answer 35

see if it makes sense

Answer 36

so in step 3, where there is x log (4) + (4) is that from distributing the 4?

Answer 37

thats from distributing log(4)

Answer 38

so then in step 4 what happend to the x that was beside the x log (4)

Answer 39

ok two steps are missing there between step 3 and step 4
3.a ) subtract xlog4 both sides
3.b ) factor the common term x in LHS

Answer 40

oh okay! that makes sense

Answer 41

great ! try second one

Answer 42

kk

Answer 43

d) is some tricky... but you would be able to figure out... try.... good luck :)

Answer 44

b) I still cant do :(

I guess i get mixed up with steps…
the last one i cant do!!!! its hard.

Answer 45

friend!

Answer 46

ok that motivates me lol

Answer 47

\(2^{2x+1} = 3^{x-2}\)

Answer 48

can i try make u do this ?

Answer 49

Ive done it!! I would take ln to both side then i get :


3x+1 ln 2 = x-2 ln 3

so then i divided by 2ln both sides and get a different answer!

Answer 50

thats nice... you understanding logarithms :)

Answer 51

I get it mostly just get caught up sometimes, gotta give me credit here!

Answer 52

i see where u getting caught up..

till taking log you are doing it right :)
once you bring the exponent to side, you should put () around it and then distribute

like this

(2x+1) ln 2 = (x-2) ln 3

Answer 53

ohhhhhhhhhhhhh

Answer 54

can u distribute nw

Answer 55

so is it 4x ln = 3xln - 6ln?

Answer 56

Noppppppee, didnt get the right answer!

Answer 57

nope let me show u


(2x+1) ln 2 = (x-2) ln 3

2x. ln2 + 1. ln2 = x.ln3 -2.ln3

Answer 58

that "2x" cannot go inside the ln. it has to stay outside lol

Answer 59

2x. ln2 + 1. ln2 = x.ln3 -2.ln3

subtracting xln3 both side
2x.ln2 - xln3 + 1. ln2 = -2.ln3

subtracting ln2 both sides
2x.ln2 - xln3 = -2.ln3 - ln2

factoring out x
x(2ln2 - ln3) = -2ln3 - ln2

x = ?

Answer 60

whats the 2x. why is there .

Answer 61

where ? first step ? its there in question itself right

Answer 62

okay i see nevermind

Answer 63

next!

Answer 64

cool.. next is ur turn lol

Answer 65

I neeed help thats why i am here! trust me ive tried all of these multiple times or else i wouldnt be on here wasting time bugging people to do it for me

Answer 66

lol its not bugging. you are learning stuff here :))

c) 50/1+ e^-x = 4

\(\huge \frac{50}{1+e^{-x}} = 4\)

its like this ?

Answer 67

yeah

Answer 68

solutiong is not that straight forward i agree.. .

Answer 69

multiply both sides with 1+ e^-x

Answer 70

\(50 = 4(1+e^{-x})\)
\(50 = 4+4e^{-x}\)
\(46 = 4e^{-x}\)
\(46/4 = e^{-x}\)

taking ln both sides

ln(46/4) = -x

x = -ln(46/4)

Answer 71

it makes sense ?

Answer 72

yes thank you sorry was doing something else again

Answer 73

its okay

Answer 74

last question can u try