Question

Can somebody help at all with:

The Human Memory Models approximate the percentage of information the average person can recall after a certain period of months have passed.
I(t) = 75-6ln(t+1)
How much does a person retain initially?
How much after 24 months?
How long does a person retain at least 60%

Please help have assignment due

Answer 1

HI!!

Answer 2

looks like no one wants to do this huh?

Answer 3

\[f(t)=75-6\ln(t+1)\]
How much does a person retain initially?
put \(t=0\) and get
\[f(0)=75-6\ln(0+1)=75\] since \(\ln(1)=0\)

Answer 4

oops
\[f(24)=75-6\ln(25)\] looks better
still need a calculator

Answer 5

ok have that one, what about the final one would it be
60=75-6ln(t+1)

Answer 6

solve most of it using algebra

Answer 7

\[-15=-6\ln(t+1)\] is the first step, then \[\frac{5}{3}=\ln(t+1)\]

Answer 8

that makes \[t+1=e^{\frac{5}{3}}\]

Answer 9

subtract one and you are done

Answer 10

have I put 60% in the correct place of the equation?

Answer 11

yes the function you are told is the percent, so your equation was correct

Answer 12

Can I ask you something else please

Answer 13

Use logarithm laws to re-write R=2/3logbase10(K)-0.9 is the formula for the richter scale and I am supposed to rewrite in the for R=logbase10 (aK^m) where a and m are constants

Answer 14

can you use the equation tool? or just rewrite it without writing "base ten" ?

Answer 15

equation tool?

Answer 16

let me try
\[\frac{2}{3}\log(K^{-0.9})\]?

Answer 17

is that is? or something else?

Answer 18

LOL I have no idea, where would the a be?

Answer 19

right below where the little blue box says \(\Sigma\)Equation

Answer 20

\[R=2/3\log_{10}(K)-0.9 \]

Answer 21

So I have to rewrite usings log laws to get \[R=\log_{10} (aK ^{m}) \] where a and m are constants derived in re-arrangement

Answer 22

on the number is outside
bring the coefficient inside the log as an exponent

Answer 23

\[\log(K^{\frac{2}{3}})\]

Answer 24

ooh you want to bring the \(0.9\) inside as well right?

Answer 25

\[R=2/3K ^{-0.9}\]

Answer 26

does that look right?

Answer 27

no

Answer 28

oh

Answer 29

\[\frac{2}{3}\log(K)-0.9\] we want to write this as a single log

Answer 30

start with
\[\frac{2}{3}\log(k)-\log(10^{-0.9})\]

Answer 31

the use the exponent property of the log and get
\[\log(K^{\frac{2}{3}})-\log(10^{-0.9})\]

Answer 32

and finally use the division property to finish it as
\[\log(\frac{K^{\frac{2}{3}}}{10^{-0.9}})\]

Answer 33

or better still
\[\huge \log(10^{0.9}K^{\frac{2}{3}})\]

Answer 34

can I ask why you minus the log10^-0.9

Answer 35

because your goal was to bring the number \(-.9\) inside the log

Answer 36

you have to use
\[\log(A)-\log(B)=\log(\frac{A}{B})\]so you need to write the number \(-0.9\) as a logarithm

Answer 37

\[\log_{10} ^{-0.9}\]

Answer 38

yeah that one

Answer 39

or \[\log10^{-0.9}\]

Answer 40

\[-0.9=\log(10^{-0.9})\]

Answer 41

You are a treasure, thank you so very very much xoxo

Answer 42

\[\color\magenta\heartsuit\]