S(t) = 68 − 20 log (t + 1), t ≥ 0 What was the average score after 4 months? after 24 months? After what time t was the average score 50%?

- anonymous

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- anonymous

Round to the nearst two decimal places

- anonymous

calculator exercise. replace t by 4, and 24. i will write it if you like

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## More answers

- anonymous

is this log base ten?

- anonymous

i dont get it

- anonymous

t = time in months yes?

- anonymous

so for the first one put
\[S(4)=68-20log(4+1)=68-20log(5)\]

- anonymous

in other words put 4 where you see a t and then compute

- anonymous

you need a calculator to find log(5)

- anonymous

i got 54.02

- anonymous

let me try

- anonymous

\[S(24)=68-20log(25)=...\]

- anonymous

i let me know if you have an answer

- anonymous

okay me to so for 24 months its
40.04

- anonymous

sound right

- anonymous

hold on i will compute

- anonymous

yup that is what i got

- anonymous

now we have to find when S = 50. do you know how to do that one?

- anonymous

no

- anonymous

first replace S(t) by 50
\[50=68-20log(t+1)\]
then get the log by itself on one side
\[20log(t+1)=68-50=18\]

- anonymous

ok so far?

- anonymous

no , why did you put 68-50

- anonymous

ok let me go step by step. \[50=68-20log(t+1)\]
add \[20log(t+1)\] to both sides to get
\[20log(t+1)+50=68\]
then subtract 50 from both sides to get
\[20log(t+1) = 68-50=18\] then divide by 20 to get
\[log(t+1)=\frac{18}{20}=\frac{9}{10}=.9\]

- anonymous

i did not need to add log(t+1) to both sides, but i prefer dealing with positive things when i am trying to solve. i could have subtracted 68 from both sides and gotten the same answer working with negative numbers. matter of preference.

- anonymous

so we have arrived via some algebra at
\[log(t+1)=.9\] is that ok so far?

- anonymous

yes i think im getting it

- anonymous

these were just algebra steps to get log(t+1) by itself. same steps i would have used to solve for x if the equation was
\[50=68-20x\]

- anonymous

you would get \[x=.9\] and here \[x=log(t+1)\] so \[log(t+1)=.9\]

- anonymous

we are not done yet, though because we have to solve for t.

- anonymous

okay im following you

- anonymous

\[log(x)=y \]
\[10^y=x\]

- anonymous

what i meant to say was
\[log(x)=y\] means the same thing as
\[10^y=x
\]

- anonymous

here we have \[log(t+1)=.9\]
so
\[10^{.9}=t+1\]

- anonymous

to find
\[10^{.9}\] in need a calculator. then to get t by itself subtract 1.

- anonymous

let me know what you get.

- anonymous

6.94 sound right

- anonymous

its wrong???

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