anonymous
  • anonymous
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%?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Owlfred
  • Owlfred
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anonymous
  • anonymous
Round to the nearst two decimal places
anonymous
  • anonymous
calculator exercise. replace t by 4, and 24. i will write it if you like

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anonymous
  • anonymous
is this log base ten?
anonymous
  • anonymous
i dont get it
anonymous
  • anonymous
t = time in months yes?
anonymous
  • anonymous
so for the first one put \[S(4)=68-20log(4+1)=68-20log(5)\]
anonymous
  • anonymous
in other words put 4 where you see a t and then compute
anonymous
  • anonymous
you need a calculator to find log(5)
anonymous
  • anonymous
i got 54.02
anonymous
  • anonymous
let me try
anonymous
  • anonymous
\[S(24)=68-20log(25)=...\]
anonymous
  • anonymous
i let me know if you have an answer
anonymous
  • anonymous
okay me to so for 24 months its 40.04
anonymous
  • anonymous
sound right
anonymous
  • anonymous
hold on i will compute
anonymous
  • anonymous
yup that is what i got
anonymous
  • anonymous
now we have to find when S = 50. do you know how to do that one?
anonymous
  • anonymous
no
anonymous
  • 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
  • anonymous
ok so far?
anonymous
  • anonymous
no , why did you put 68-50
anonymous
  • 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
  • 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
  • anonymous
so we have arrived via some algebra at \[log(t+1)=.9\] is that ok so far?
anonymous
  • anonymous
yes i think im getting it
anonymous
  • 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
  • anonymous
you would get \[x=.9\] and here \[x=log(t+1)\] so \[log(t+1)=.9\]
anonymous
  • anonymous
we are not done yet, though because we have to solve for t.
anonymous
  • anonymous
okay im following you
anonymous
  • anonymous
\[log(x)=y \] \[10^y=x\]
anonymous
  • anonymous
what i meant to say was \[log(x)=y\] means the same thing as \[10^y=x \]
anonymous
  • anonymous
here we have \[log(t+1)=.9\] so \[10^{.9}=t+1\]
anonymous
  • anonymous
to find \[10^{.9}\] in need a calculator. then to get t by itself subtract 1.
anonymous
  • anonymous
let me know what you get.
anonymous
  • anonymous
6.94 sound right
anonymous
  • anonymous
its wrong???

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