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jagatuba

  • 5 years ago

Can anyone show me how to prove two equations are equivalent?

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  1. anonymous
    • 5 years ago
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    which equations?

  2. jagatuba
    • 5 years ago
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    L*[c/(1- (1 + c) ^ -n) ] and L*[c(1 + c)^n]/[(1 + c)^n - 1]

  3. anonymous
    • 5 years ago
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    ???

  4. jagatuba
    • 5 years ago
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    Bot of those expressions should equal P

  5. jagatuba
    • 5 years ago
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    In other words: P = L*[c/(1- (1 + c) ^ -n) ] and P = L*[c(1 + c)n]/[(1 + c)n - 1]

  6. anonymous
    • 5 years ago
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    L , c, n are variables ???

  7. jagatuba
    • 5 years ago
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    Yes I get as far as eliminating L by division, but then I get stuck.

  8. anonymous
    • 5 years ago
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    Could you use the equation button to express it in a better way....I m pretty gud at solving equations...but its just written weirdly...

  9. jagatuba
    • 5 years ago
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    Yeah I know what you mean. I'll try. I was messing with the equation button last night but was not getting the equations to look right. lol Give me a few

  10. anonymous
    • 5 years ago
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    ok

  11. jagatuba
    • 5 years ago
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    \[P=L \times \left[ c \div \left( 1-\left( 1+c \right)^{-n} \right) \right]\] \[P=L \times \left[ \left( c \times \left( 1+c \right)^{n} \right) \div \left( \left( 1+c \right)^{n} \right)-1\right]\]

  12. jagatuba
    • 5 years ago
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    Sorry it took so long

  13. anonymous
    • 5 years ago
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    no prob...let me try

  14. anonymous
    • 5 years ago
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    heyyyyyyyy

  15. anonymous
    • 5 years ago
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    so simple ...it was...

  16. jagatuba
    • 5 years ago
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    I don't know if this will help you at all but these are two equations that are used to determine the monthly payment on a mortgage loan (P=monthly payment). I know they do come up with the same results I just have to prove mathematically

  17. anonymous
    • 5 years ago
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    in the first equation, make (1+c)^ -n = 1/ (1+c)^ n and then take the lcm

  18. jagatuba
    • 5 years ago
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    (1=c)?

  19. jagatuba
    • 5 years ago
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    'Scuse me (1+c)?

  20. anonymous
    • 5 years ago
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    yes in the first equation...

  21. jagatuba
    • 5 years ago
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    I could write the right side as 1/(c+1)(c+1) but how would I write the left side with it's negative exponent?

  22. anonymous
    • 5 years ago
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    but that was (1+c)^n ...wasnt it?

  23. jagatuba
    • 5 years ago
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    -n is the exponent

  24. jagatuba
    • 5 years ago
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    Wait now I'm getting confused. lol

  25. anonymous
    • 5 years ago
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    OK...let me make sure again....we are trying to prove that first equation equals to second equation...right??

  26. anonymous
    • 5 years ago
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    yes...I was right...

  27. anonymous
    • 5 years ago
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    in the first equation...take the lcm in the denominator

  28. jagatuba
    • 5 years ago
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    Okay give me a minute to work this out on paper and I'll post again.

  29. anonymous
    • 5 years ago
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    which will eventually go to numerator

  30. jagatuba
    • 5 years ago
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    SO after dividing both sides of the equation by L I get: \[\left( c \div \left( 1-\left( 1+c \right)^{-n} \right) \right)=\left( \left( c \times \left( 1+c \right)^{n} \right)\div \left( \left( 1+c \right)^{n} \right)-1 \right)\] Next I take care of the negative exponent like this: \[c-\left( 1+c \right)^{n }=\left( c \times \left( 1+c \right)^{n} \right)\div \left( \left( 1+c \right)^{n}-1 \right)\] Is this right so far?

  31. jagatuba
    • 5 years ago
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    Actually that right side should be \[c \times -\left( 1+c \right)^{n}\]

  32. jagatuba
    • 5 years ago
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    But now I'm lost again. Dang it.

  33. anonymous
    • 5 years ago
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    u have got it !!!

  34. jagatuba
    • 5 years ago
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    But if I divide each side by \[-\left( 1+c \right)^{n}\] I still end up with: \[c =c \div \left( \left( 1+c \right)^{n}-1 \right)\]

  35. jagatuba
    • 5 years ago
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    That can't be right can it?

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