SOLUTION: Students in an English class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. The average score S(t), in percent, after t months was found
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Question 439661: Students in an English class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. The average score S(t), in percent, after t months was found to be given by S(t) = 68 -20 log (t + 1), t ≥ 0.
After what time t was the average score 50%?
This is all I got so far... I do not know where to start or finish please help! thank you.
.5= 68 - 20 log (t + 1), t ≥ 0.
Found 2 solutions by stanbon, tpcrayton:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Students in an English class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. The average score S(t), in percent, after t months was found to be given by S(t) = 68 -20 log (t + 1), t ≥ 0.
After what time t was the average score 50%?
This is all I got so far... I do not know where to start or finish please help! thank you.
50 = 68 - 20 log (t + 1), t ≥ 0.
----
-18 = -20*log(t+1)
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0.9 = log(t+1)
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t+1 = 10^0.9
---
t = 7.943.. -1
---
t = 6.943 months
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Cheers,
Stan H.
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Answer by tpcrayton(1) (Show Source): You can put this solution on YOUR website!
50=68-20log(t+1)
-18 = -20log(t+1)
-18/-20=-20/-20*log(t+1)
9/10 = log10(t+1)
t+1=10^0.9
t+1-1=7.94-1
t= 6.94
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