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anonymous
 5 years ago
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
 5 years ago
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%?

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.1Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Round to the nearst two decimal places

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0calculator exercise. replace t by 4, and 24. i will write it if you like

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this log base ten?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0t = time in months yes?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so for the first one put \[S(4)=6820log(4+1)=6820log(5)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in other words put 4 where you see a t and then compute

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need a calculator to find log(5)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[S(24)=6820log(25)=...\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i let me know if you have an answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay me to so for 24 months its 40.04

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hold on i will compute

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yup that is what i got

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we have to find when S = 50. do you know how to do that one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first replace S(t) by 50 \[50=6820log(t+1)\] then get the log by itself on one side \[20log(t+1)=6850=18\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no , why did you put 6850

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok let me go step by step. \[50=6820log(t+1)\] add \[20log(t+1)\] to both sides to get \[20log(t+1)+50=68\] then subtract 50 from both sides to get \[20log(t+1) = 6850=18\] then divide by 20 to get \[log(t+1)=\frac{18}{20}=\frac{9}{10}=.9\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0so we have arrived via some algebra at \[log(t+1)=.9\] is that ok so far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i think im getting it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0these 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 \[50=6820x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you would get \[x=.9\] and here \[x=log(t+1)\] so \[log(t+1)=.9\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we are not done yet, though because we have to solve for t.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay im following you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[log(x)=y \] \[10^y=x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what i meant to say was \[log(x)=y\] means the same thing as \[10^y=x \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here we have \[log(t+1)=.9\] so \[10^{.9}=t+1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to find \[10^{.9}\] in need a calculator. then to get t by itself subtract 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me know what you get.
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