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

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

Mathematics
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which equations?
L*[c/(1- (1 + c) ^ -n) ] and L*[c(1 + c)^n]/[(1 + c)^n - 1]
???

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Bot of those expressions should equal P
In other words: P = L*[c/(1- (1 + c) ^ -n) ] and P = L*[c(1 + c)n]/[(1 + c)n - 1]
L , c, n are variables ???
Yes I get as far as eliminating L by division, but then I get stuck.
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...
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
ok
\[P=L \times \left[ c \div \left( 1-\left( 1+c \right)^{-n} \right) \right]\] \[P=L \times \left[ \left( c \times \left( 1+c \right)^{n} \right) \div \left( \left( 1+c \right)^{n} \right)-1\right]\]
Sorry it took so long
no prob...let me try
heyyyyyyyy
so simple ...it was...
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
in the first equation, make (1+c)^ -n = 1/ (1+c)^ n and then take the lcm
(1=c)?
'Scuse me (1+c)?
yes in the first equation...
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?
but that was (1+c)^n ...wasnt it?
-n is the exponent
Wait now I'm getting confused. lol
OK...let me make sure again....we are trying to prove that first equation equals to second equation...right??
yes...I was right...
in the first equation...take the lcm in the denominator
Okay give me a minute to work this out on paper and I'll post again.
which will eventually go to numerator
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?
Actually that right side should be \[c \times -\left( 1+c \right)^{n}\]
But now I'm lost again. Dang it.
u have got it !!!
But if I divide each side by \[-\left( 1+c \right)^{n}\] I still end up with: \[c =c \div \left( \left( 1+c \right)^{n}-1 \right)\]
That can't be right can it?

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