Question

The logistic growth function: f(t)=91965/(1+1513e^-t)

What is the limiting size of the population that becomes ill?

Answer 1

you are being asked for the limit as \(t\to \infty\)

Answer 2

Knowing that at the beginning, 18 people became ill, and after 2 weeks about 132 people became ill

Answer 3

Wow you really have my back tonight thank you hahaha

Answer 4

it is just the number in the numerator

Answer 5

as \(t\to \infty\) you have \(e^{-t}\to 0\)

Answer 6

so the denominator gets close to 1 and the whole thing gets close to 91965

Answer 7

Okay perfect. Thank you very much

Answer 8

this stuff
Knowing that at the beginning, 18 people became ill, and after 2 weeks about 132 people became ill
is just extra verbiage to set up the problem

yw

Answer 9

Am I just suppose to assume that e^-t=0?

Answer 10

in the limit
as t gets larger \(e^{-t}=\frac{1}{e^t}\) goes to zero very very fast

Answer 11

think of it this way
\[e^t\] grows very quickly
that means
\[\frac{1}{e^t}\] has a very big denominator for large values of \(t\)
big denominator means the number is very close to zero

Answer 12

ps don't forget \(e^{-t}\) really means \(\frac{1}{e^t}\)

Answer 13

right! awesome thanks

Answer 14

yw

Answer 15

while youre already here...

Answer 16

yes...

Answer 17

how would I write this equation in irs equivilent exponential form

logv14185=y

Answer 18

that v is my way of saying like an underscore

Answer 19

let me guess
is it
\[\log_{14}(185)=y\]?

Answer 20

so log(lower exponent 14)185

Answer 21

yes hahah

Answer 22

use underscore it works
in any case what you need to know is this
\[b^y=x\iff \log_b(x)=y\] and you should be able to switch quickly end effortlessly
for example \[2^3=8\iff \log_2(8)=3\] and
\[10^{-2}=.01\iff \log_{10}(.01)=-2\]

Answer 23

ooooo I forgot about that property

Answer 24

so
\[\log_{14}(185)=y\iff 14^y=185\]

Answer 25

they say the same thing, and that is how you are going to solve logarithmic and exponential equations
that, plus the change of base formula

Answer 26

Okay so what would the b, x, and y be for ^5sqrt776=6

Answer 27

do you need to simplify it first before you determine that?

Answer 28

Or can you tell just by looking immediately

Answer 29

i can't really figure out what you wrote
give me a second

Answer 30

I might have written it wrong

Answer 31

\[\sqrt[5]{776}=6\]

Answer 32

yes lol

Answer 33

is that right?

Answer 34

yes thanks

Answer 35

i am not sure what you are supposed to do with it
it is an equality between numbers
you can write it as an exponent though
lets see if that is what is needed

Answer 36

\[\large \sqrt[5]{776}=776^{\frac{1}{5}}\]

Answer 37

I am asked to give the base, x, and y of that equation

Answer 38

since
\[\large 776^{\frac{1}{5}} =6\] we can also say
\[\log_{776}(6)=\frac{1}{5}\]

Answer 39

so one thing you could say is that the base is \(b=776\)
not sure what the \(x\) or \(y\) are supposed to be
it is
\[\log_b(x)=y\]? in that case \[b=776, x=6,y=\frac{1}{5}\]

Answer 40

thank you so much you are seriously saving my life tonight lol

Answer 41

lol
not sure if i saved your life, but glad to help

Answer 42

So if I evaluate log_7(1/^3sqrt7) how would I do that

Answer 43

so far I have 7^x= 1/(^3sqrt7)

Answer 44

by sqrt I just mean the symbol. so the cube root of 7 i guess

Answer 45

or I guess the other way to write that would be (1/(7^1/3))

Answer 46

so now I have 7^x=1/(7^(1/3))

Answer 47

\[7^x=\frac{1}{\sqrt[3]{7}}\]

Answer 48

yes

Answer 49

which you wrote as
\[7^x=\frac{1}{7^{\frac{1}{3}}}\] but it would help your cause more if you wrote it as
\[7^x=7^{-\frac{1}{3}}\]

Answer 50

You can do that??

Answer 51

then you would immediately see that \(x=-\frac{1}{3}\)

Answer 52

yes of course

Answer 53

\[b^{-n}=\frac{1}{b^n}\]

Answer 54

Ah I guess that makes sense

Answer 55

so \[\large \frac{1}{7^{\frac{1}{3}}}=7^{-\frac{1}{3}}\]

Answer 56

it is not really a matter of making sense, it is the way you use exponential notation
since logarithms are exponents (at least at the introductory level) it is all about knowing exponential notation

Answer 57

okay i see

Answer 58

What if I need to evaluate it when it is all in the exponent? Like: 2^(log_2)69

Answer 59

knowing for example that
\[\sqrt[n]{x^m}=x^{\frac{m}{n}}\] and stuff like that

Answer 60

2^((log_2)69)

Answer 61

you have
\[\large 2^{\log_2(69)}\]?

Answer 62

I have 2^x=69 but I still have that other 2 to deal with

Answer 63

yes

Answer 64

ok lets go slow because this one is very very easy

Answer 65

ok sweet haha

Answer 66

this is not an answer to your question, it is my example to make it clear

Answer 67

\[2^4=16\] which means
\[\log_2(16)=4\]
so what is
\[2^{\log_2(16)}\]? well it must be \(2^4\) which is \(16\)
in other words it is always true that
\[\huge 2^{\log_2(\diamondsuit)}=\diamondsuit\]

Answer 68

the log is an exponent.
\(\log_2(69)\) is the the number you raise two to in order to get \(69\)which makes
\[\large 2^{\log_2(69)}=69\]

Answer 69

more generally
\[\huge b^{\log_b(x)}=x\]

Answer 70

so 69?

Answer 71

for example
\[e^{\ln(7)}=7\]

Answer 72

yes so 69 easy as that

Answer 73

another way to say this is that the log is the inverse of the exponential

Answer 74

so
\[\log_b(b^x)=x\] and
\[b^{\log_b(x)}=x\]

Answer 75

thank you :) How would I write h(x)=log_5(x)+1

Answer 76

I am now given multiple equations like these to match up with specific graphs

Answer 77

i am not sure what you mean "how would i write it"

Answer 78

\[h(x)=\log_5(x)+1\] looks like a basic log function moved up one unit

Answer 79

How would I rewrite it I guess.. to graph on a calc. since I cant technically write that in a calculator like that

Answer 80

oooh i see
here take a look at this
http://www.wolframalpha.com/input/?i=log_5%28x%29%3D1+domain+.1..20

Answer 81

you can in fact use a graphing calculator although i would use wolfram
i can show you how if you like

Answer 82

okay can you show me how? And why did you set that equal to 1

Answer 83

oh damn i put equal one, not plus one
my mistake

Answer 84

http://www.wolframalpha.com/input/?i=log_5%28x%29%2B1+domain+.1..20

Answer 85

I'm thinking for testing purposes I wont have wolfram unfortunately haha

Answer 86

but that is def helpful for homework

Answer 87

ok well here is what you can do for a graphing calculator
it is always the case that
\[\log_b(x)=\frac{\ln(x)}{\ln(b)}\] so for example
\[\log_5(x)+1=\frac{\ln(x)}{\ln(5)}+1\]

Answer 88

that log you do have in your graphing calculator so you can use it if you like
but if you have a test it might be hard to match up what you calculator shows you with what is on the test paper so you have to be a bit savvy on how you use it, or else just know something about the graph to begin with

Answer 89

I'm wolframming these for now and some arent matching up perfectly so I'm just going with what is closest to what it looks like

Answer 90

you want to post a screen shot or something? i am sure there will be clues to figure it out
it is tough with graphs because they all look different
you are probably supposed to figure it out some other way

Answer 91

one clue is where they cross the x or y axis
although even that might not be enough

Answer 92

Yeah okay one sec ill post pics

Answer 93

A)

Answer 94

B)

Answer 95

C)

Answer 96

before we start
is there an equation or are you supposed to match them up

Answer 97

D)