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

- jagatuba

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

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- anonymous

which equations?

- jagatuba

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

- anonymous

???

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## More answers

- jagatuba

Bot of those expressions should equal P

- jagatuba

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

- anonymous

L , c, n are variables ???

- jagatuba

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

- 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...

- 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

- anonymous

ok

- jagatuba

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

- jagatuba

Sorry it took so long

- anonymous

no prob...let me try

- anonymous

heyyyyyyyy

- anonymous

so simple ...it was...

- 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

- anonymous

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

- jagatuba

(1=c)?

- jagatuba

'Scuse me (1+c)?

- anonymous

yes in the first equation...

- 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?

- anonymous

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

- jagatuba

-n is the exponent

- jagatuba

Wait now I'm getting confused.
lol

- anonymous

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

- anonymous

yes...I was right...

- anonymous

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

- jagatuba

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

- anonymous

which will eventually go to numerator

- 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?

- jagatuba

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

- jagatuba

But now I'm lost again. Dang it.

- anonymous

u have got it !!!

- jagatuba

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

- jagatuba

That can't be right can it?

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