## 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%?

1. Owlfred

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2. anonymous

Round to the nearst two decimal places

3. anonymous

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

4. anonymous

is this log base ten?

5. anonymous

i dont get it

6. anonymous

t = time in months yes?

7. anonymous

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

8. anonymous

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

9. anonymous

you need a calculator to find log(5)

10. anonymous

i got 54.02

11. anonymous

let me try

12. anonymous

$S(24)=68-20log(25)=...$

13. anonymous

i let me know if you have an answer

14. anonymous

okay me to so for 24 months its 40.04

15. anonymous

sound right

16. anonymous

hold on i will compute

17. anonymous

yup that is what i got

18. anonymous

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

19. anonymous

no

20. anonymous

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

21. anonymous

ok so far?

22. anonymous

no , why did you put 68-50

23. 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 $log(t+1)=\frac{18}{20}=\frac{9}{10}=.9$

24. 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.

25. anonymous

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

26. anonymous

yes i think im getting it

27. 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 $50=68-20x$

28. anonymous

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

29. anonymous

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

30. anonymous

okay im following you

31. anonymous

$log(x)=y$ $10^y=x$

32. anonymous

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

33. anonymous

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

34. anonymous

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

35. anonymous

let me know what you get.

36. anonymous

6.94 sound right

37. anonymous

its wrong???