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The Bonus Question for Episode 3 of Double Oh 3.14!

The Bonus Question for Episode 3 of Double Oh 3.14!

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  • 1. Bonus Question! Is this really necessary?
  • 2. Question
    • Ok, so that wasn’t the real question. It seems like Agent Double Oh 3.14 has found another clue after beating the Hooded Bandito. And the chase and fight drained most of her energy! She only has enough energy to use her calculator once. But what’s this clue about? Let’s find out!
    • There are words written on the paper.
    • Find the sum of the first (e ln 7 )(log 4 64 3 ) terms in an arithmetic sequence whose first term is: And whose second term is:
    • 3(log 3 ( 9 C 8 • 3 C 1 ) – log 3 ( ))
  • 3. DO NOT MOVE ON UNTIL YOU HAVE ANSWERED THE QUESTION OR YOU NEED HELP!
  • 4. Things You Should Know
    • Alright. This question has content from 3 different units: Combinatorics, Logarithms and Exponents and Sequences.
    • Also, we can only use our calculators ONCE (and be honest with yourselves)!
    • So let’s start by decoding the message:
    • Find the sum of the first (e ln 7 )(log 4 64 3 ) terms……
  • 5. Decoding the Message (Logs and Exp.)
    • (e ln 7 )(log 4 64 3 )
    • Let’s look at this. Do we need our calculators?
    • NO WAY!!
    • For the first set of brackets, e ln 7 , the answer is 7.
    • Why?
    • BECAUSE A LOGARITHM IS AN EXPONENT (and I bet you forgot that!)
    • “ ln 7” means “What exponent on base “e” will give you 7?” and once you’ve found that exponent, you are putting it on base “e” therefore the answer is 7.
  • 6. Decoding the Message (Logs and Exp.)
    • Confused? No worries, let me explain.
    • Ex. 10 log 1000
    • What does this mean?
    • “ 10 to the power of “an exponent that you put on base 10 that equals 1000””
    • “ An exponent that you put on base 10 that equals 1000” is 3.
    • When you rewrite it you put it on a base of 10 and you get 10 3 which equals 1000.
    • We can conclude by saying:
  • 7. Decoding the Message (Logs and Exp.)
    • (e ln 7 )(log 4 64 3 )
    • Alright now the second set of brackets.
    • You can use the power law on this expression.
    • The power law simply goes like this:
    • log a M N =N log a M
    • So log 4 64 3 = 3 log 4 64.
    • log 4 64 you should be able to do in your head. log 4 64=3
    • log 4 64 3 = 3 log 4 64 = 3(3) = 9
  • 8. Decoding the Message (Logs and Exp.)
    • (e ln 7 )(log 4 64 3 ) = (7)(9) = 63
    • We have just decoded the first part of the message!
    • Find the sum of the first 63 terms……
    • Now the next part:
    • … in an arithmetic sequence whose first term is:
  • 9. Decoding the Message (Logs and Exp.)
    • There are two ways to approach this. One way is to do each logarithmic expression in your head.
    • log 2 64 = 6 log 2 8 = 3 6/3= 2
    • =2
  • 10. Decoding the Message (Logs and Exp.)
    • The second way is to recognize that the change of base law has been applied here. The change of base law goes like this:
    • So = log 8 64
    • This you can do in your head.
    • log 8 64 = 2
  • 11. Decoding the Message (Logs and Exp.)
    • Hooray! We got the second part!
    • Find the sum of the first 63 terms in an arithmetic sequence whose first term is 2…
    • Now the third part of the message.
    • … And whose second term is:
    • 3(log 3 ( 9 C 8 • 3 C 1 ) – log 3 ( ))
    • Alright! Let’s decode this!!!
  • 12. Decoding the Message (Logs and Exp.)
    • 3(log 3 ( 9 C 8 • 3 C 1 ) – log 3 ( ))
    • So let’s start with the Combinatorics in the logarithm expressions.
    • Do we need a calculator? NO!
    • As long as you remember that the Choose formula is:
    • n C r =
    • And the Pick formula is:
    • n P r =
  • 13. Decoding the Message (Logs and Exp.)
    • 3(log 3 ( 9 C 8 • 3 C 1 ) – log 3 ( ))
    • Let’s start with the beginning:
    • 9 C 8 = = = = =9
    • 3 C 1 = = = = =3
    • 9 C 8 • 3 C 1 =9*3=27
  • 14. Decoding the Message (Logs and Exp.)
    • 3(log 3 (27) – log 3 ( ))
    • Now the second part:
    • 6 P 2 = = = = (6)(5)=30
    • 5 C 3 = = = = = =10
    • =
  • 15. Decoding the Message (Logs and Exp.)
    • 3(log 3 (27) – log 3 (3))
    • Now we solve it! And no, we don’t need a calculator.
    • 3(log 3 (27) – log 3 (3))
    • =3(3 – 1)
    • =3(2)
    • =6
    • Now we’ve got the full message!
    • Find the sum of the first 63 terms in an arithmetic sequence whose first term is 2 and whose second term is 6.
    • Now we have to solve this question!
  • 16. Decoding the Message (Logs and Exp.)
    • Find the sum of the first 63 terms in an arithmetic sequence whose first term is 2 and whose second term is 6.
    • To find the sum of an arithmetic sequence, we use the formula:
    • This means we need to find:
    • >The first term.
    • >The number of terms (n) we are finding the sum of.
    • >The ‘n’th term.
    • Well we know the first term is 2, the number of terms we are finding the sum of is 63, but we don’t know the 63 rd term.
    • To find it, we use this formula:
  • 17. Decoding the Message (Logs and Exp.)
    • To use this formula, we need to know:
    • >The number of the term we are looking for. (n)
    • >The first term. (t 1 )
    • >The common difference. (d)
    • We know we are looking for the 63 rd term, the first term is 2, but we don’t know the common difference.
    • To find it, we just subtract one term from the term directly after it.
    • We know t 1 =2 and the t 2 =6 so the common difference is t 2 -t 1 =6-2 =4
    • Now we plug it into the equation to find the 63 rd term.
    • t 63 =2+4(63-1)
    • t 63 =2+4(62)
    • t 63 =2+248
    • t 63 =250 Now that we know the 63 rd term, we can plug it into the sum equation.
  • 18. Decoding the Message (Logs and Exp.)
    • n=63
    • t 1 =2
    • t 63 =250
    • Now would be the time to use your calculator unless you’re fine doing this in your head.
    • S 63 =63/2(2+250)
    • S 63 =63/2(252)
    • S 63 =7938
  • 19. So the sum of the terms is 7938, but what does that mean?
  • 20. Highlight under “BONUS QUESTION TIME!!” to move on to episode 4.