Question

1 question, will fan!

Answer 1

What is that one question...

Answer 2

@satellite73 could you help again, i'm trying to get things for my study guide right.

Answer 3

\[f(5)=5500\ln(9\times 5+4)=5500\ln(49)\] and a calculator

Answer 4

in plain english, where you see a \(t\) replace it by a \(5\)

Answer 5

i got 275000

Answer 6

let me check, seems large

Answer 7

5500 times 49

Answer 8

oh no dear, the log of 49, not 49

Answer 9

Thats what ive been doing wrong then

Answer 10

you forgot the \(\ln(49)\) part

Answer 11

i get \(21405\) rounded
http://www.wolframalpha.com/input/?i=5500ln%2849%29

Answer 12

that make more sense?

Answer 13

Ok, so what exactly is In? thats whats throwing me off.

Answer 14

\(\ln(x)\) is a function
you plug in a number, get out a number
you usually have to use a calculator to evaluate it, so if i want \(\ln(49)\) like in this problem, you need a calculator

Answer 15

Ok, thanks!, i only have a few more, think you could help?

Answer 16

it is the inverse function of the exponential function \(f(x)=e^x\) in other words if
\[\ln(x)=y\iff x=e^y\] you should be able to switch back and forth easily
just like \(\log_2(x)=y\iff x=2^y\)
sure shoot

Answer 17

i mean "sure, go ahead and ask, i will answer or help"

Answer 18

to increase a number by \(11\%\) multiply it by \(1.11\) as \(11\%=.11\)

Answer 19

if you start with \(330\) deer, after 1 year there will be
\[330\times 1.11\] deer
after two year there will be
\[330\times 1.11\times 1.11=330(1.11)^2\] deer, and after \(t\) years there will be
\[330\times (1.11)^t\] deer

Answer 20

ok, following so far.

Answer 21

\[\large \color{red}{330}\times( \color{blue}{1.11})^{\color{green}t}\]
\(\color{red}{330}\) is the original population of deer, what you have when you start counting

Answer 22

\(\color{blue}{1.11}\) is \(1+.11\) how you increase a number by \(11\%\) and
\(\color{green}t\) is the number of years after you start counting

Answer 23

and so after 5 years you expect to have
\[330\times (1.11)^5\] deer
you need a calculator to compute this number

Answer 24

i got 606.620936

Answer 25

now not to confuse you, but if it is \(11\%\) compounded continuously and not "per year" which is what it says, then you would use
\[\large 330\times e^{.11t}\]

Answer 26

i get \(556\)
http://www.wolframalpha.com/input/?i=330%281.11%29^5

Answer 27

maybe you made a typo when you put it in your calculator
i like wolfram cause you just type it in as you see it

Answer 28

I am going to have to start using that! haha

Answer 29

yeah, good idea

Answer 30

Ok so after 5 years it will be 556?

Answer 31

that is what i get, yes

Answer 32

ok two more questions and the study guide is complete! :)

Answer 33

the original price of the boat was $3500 so we know it will be \[3500\times b^x\] we just don't know \(b\) yet

Answer 34

ok so start off with 3500 times b^x

Answer 35

right

Answer 36

now lets see if we can find another good point on the graph
it looks like after 2 years it is at $2000

Answer 37

so it has decreased by a factor of \(\frac{2000}{3500}=\frac{4}{7}\) so that will be our \(b\) but don't forget that was two year not one year

Answer 38

ok

Answer 39

so unless you want to estimate the value at \(t=1\) i would use
\[3500\times \left(\frac{4}{7}\right)^{\frac{t}{2}}\]

Answer 40

ok so right now b=4/7 and x=t/2?

Answer 41

the reason for the 2 in the denominator of the exponent is because that was the decrease for two years, not one

Answer 42

right, what you said

Answer 43

to finish, compute at \(t=9.5\) and get
\[3500\times \left(\frac{4}{7}\right)^{\frac{9.5}{2}}\]

Answer 44

careful putting this in the calculator, there are lots of ways to make mistakes here

Answer 45

im going to use that one website you were using, hold on.

Answer 46

i got 8.59363

Answer 47

ok
if you use a calculator i would first compute \(9.5\div 2=4.75\) and only then compute
\[3500(\frac{4}{7})^{4.75}\]

Answer 48

careful with the syntax
send me the link of what you wrote in wolfram

Answer 49

i am saying that because i got something else

Answer 50

http://www.wolframalpha.com/input/?i=3500*%284%2F7%29%5E9.5%2F2

Answer 51

something went wrong though.

Answer 52

yeah what went wrong is you didn't put parentheses around the exponent
you can see underneath what wolfram is computing, and it is not what you want

Answer 53

http://www.wolframalpha.com/input/?i=3500*%284%2F7%29%5E%289.5%2F2%29

Answer 54

that is why i mentioned computing \(9.5\div2=4.75\) first, then you wont have that mistake

Answer 55

245.266

Answer 56

much better

Answer 57

Thats the final answer?

Answer 58

since it is money, i would round

Answer 59

\(\$245.27\) maybe

Answer 60

Ok thats what i was typing haha

Answer 61

or just \(\$245\) depending on how picky you want to be

Answer 62

I'll just do 245.27

Answer 63

k good
that it?

Answer 64

One last one, i think i know this one but im going to get your thoughts on it

Answer 65

k

Answer 66

nice softball question there

Answer 67

K is what controls to line rising and decreasing, when k is negative the line will decrease but when k is positive it will rise, since k is positive it is going to rise

Answer 68

control the line vertically*

Answer 69

okay slow here
it is not a "line" but rather a "curve"

Answer 70

ok.

Answer 71

it shift the graph of \[y=ab^{x-h}+k\] up \(k\) units if \(k>0\)and down \(k\) units if \(k<0\)
unless you are told \(k\) is positive, you cannot assume it is, because it is a variable

Answer 72

i guess i should really say "it shifts the graph of \(y=ab^{x-h}\) up or down

Answer 73

Alright, so k shifts the graph of y=ab^x-h either up or down?

Answer 74

yes, the shape of the curve remains unchanged, it just goes up or down
and don't call it a "line" or your math teacher will think you are confused

Answer 75

Thanks, i just put that is shift the graph of that equation up or down

Answer 76

k sound good
done?

Answer 77

Done, thanks so much!

Answer 78

yw
you owe me a beer

Answer 79

haha i don't know how you didn't give up.

Answer 80

just kidding

Answer 81

oh i never give up
you did very well, especially with the one about the boat
i think you got most of this, even if you are a bit confused about the log

Answer 82

I have another study guide for this class for extra credit but its a lot so idk if im going to do it tonight or not.

Answer 83

take a break

Answer 84

I am, i might complete it in a bit, but as of now im going to relax.

Answer 85

me, i am going to go to bed
later

Answer 86

Bye :) Thanks again.