## jagatuba 5 years ago Can anyone show me how to prove two equations are equivalent?

1. anonymous

which equations?

2. jagatuba

L*[c/(1- (1 + c) ^ -n) ] and L*[c(1 + c)^n]/[(1 + c)^n - 1]

3. anonymous

???

4. jagatuba

Bot of those expressions should equal P

5. jagatuba

In other words: P = L*[c/(1- (1 + c) ^ -n) ] and P = L*[c(1 + c)n]/[(1 + c)n - 1]

6. anonymous

L , c, n are variables ???

7. jagatuba

Yes I get as far as eliminating L by division, but then I get stuck.

8. anonymous

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

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

ok

11. jagatuba

$P=L \times \left[ c \div \left( 1-\left( 1+c \right)^{-n} \right) \right]$ $P=L \times \left[ \left( c \times \left( 1+c \right)^{n} \right) \div \left( \left( 1+c \right)^{n} \right)-1\right]$

12. jagatuba

Sorry it took so long

13. anonymous

no prob...let me try

14. anonymous

heyyyyyyyy

15. anonymous

so simple ...it was...

16. jagatuba

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

in the first equation, make (1+c)^ -n = 1/ (1+c)^ n and then take the lcm

18. jagatuba

(1=c)?

19. jagatuba

'Scuse me (1+c)?

20. anonymous

yes in the first equation...

21. jagatuba

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

but that was (1+c)^n ...wasnt it?

23. jagatuba

-n is the exponent

24. jagatuba

Wait now I'm getting confused. lol

25. anonymous

OK...let me make sure again....we are trying to prove that first equation equals to second equation...right??

26. anonymous

yes...I was right...

27. anonymous

in the first equation...take the lcm in the denominator

28. jagatuba

Okay give me a minute to work this out on paper and I'll post again.

29. anonymous

which will eventually go to numerator

30. jagatuba

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

Actually that right side should be $c \times -\left( 1+c \right)^{n}$

32. jagatuba

But now I'm lost again. Dang it.

33. anonymous

u have got it !!!

34. jagatuba

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

That can't be right can it?