Question

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

Answer 1

which equations?

Answer 2

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

Answer 3

???

Answer 4

Bot of those expressions should equal P

Answer 5

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

Answer 6

L , c, n are variables ???

Answer 7

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

Answer 8

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...

Answer 9

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

Answer 10

ok

Answer 11

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

Answer 12

Sorry it took so long

Answer 13

no prob...let me try

Answer 14

heyyyyyyyy

Answer 15

so simple ...it was...

Answer 16

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

Answer 17

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

Answer 18

(1=c)?

Answer 19

'Scuse me (1+c)?

Answer 20

yes in the first equation...

Answer 21

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?

Answer 22

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

Answer 23

-n is the exponent

Answer 24

Wait now I'm getting confused.
lol

Answer 25

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

Answer 26

yes...I was right...

Answer 27

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

Answer 28

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

Answer 29

which will eventually go to numerator

Answer 30

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?

Answer 31

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

Answer 32

But now I'm lost again. Dang it.

Answer 33

u have got it !!!

Answer 34

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

Answer 35

That can't be right can it?