Let X = substitution

nquadr · **Joined:** Sun, 5 Feb 2012 21:05:29 UTC **Posts:** 18

Hi can anyone help me simplify this equation by letting X = someNumber, to help me solve this equation.

24500(1+i)^3 - 10000(1+i)^(5/3) + 8000(1+i)^(13/12) = 30309

I would like to simplify it algebraically. Thanks.

Denis · **Posted:** Thu, 1 Mar 2012 02:51:28 UTC

nquadr wrote:

Hi can anyone help me simplify this equation by letting X = someNumber, to help me solve this equation.
24500(1+i)^3 - 10000(1+i)^(5/3) + 8000(1+i)^(13/12) = 30309
I would like to simplify it algebraically. Thanks.

Huh? Not much can be done; let x = 1+i; then:
x^3[16/x^(23/12) - 20/x^(4/3) + 49] = 30309/500

Iteration is required to solve; can't be solved directly;
I got x = ~1.10685, so i = ~0.10685

Your equation could represent:
Future Value of $24,500 @ 10.685% annually for 3 years
less: FV of $10,000 @ 10.685% annually for 5/3 years
plus: FV of $8,000 @ 10.685% annually for 13/12 years

If you don't follow that, then: why are you working with that equation? :idea:

Shadow · **Posted:** Thu, 1 Mar 2012 02:52:10 UTC

nquadr wrote:

Hi can anyone help me simplify this equation by letting X = someNumber, to help me solve this equation.

24500(1+i)^3 - 10000(1+i)^(5/3) + 8000(1+i)^(13/12) = 30309

I would like to simplify it algebraically. Thanks.

Write 3=36/12 and 5/3=20/12 then you get:

$(1+i)^{13/12}(24500(1+i)^{23/12}+10000(1+i)^{7/12}+8000)=30309$ , but I'm pretty sure you have little chance of getting anything out of this to solve explicitly for i using analytic methods.

nquadr · **Joined:** Sun, 5 Feb 2012 21:05:29 UTC **Posts:** 18

Yes, you are right, they are theory of interest problems. I was solving this equation with my TI-89 calculator, and it was taking a long time... I thought I could simplify further, but I guess I can't. Do you have a link to teach me the iteration method by hand? Just curious that's all. Once again, thanks for your help.

Denis · **Posted:** Thu, 1 Mar 2012 06:36:05 UTC

http://www.google.ca/#hl=en&gs_nf=1&cp= ... 04&bih=560

Soroban · **Posted:** Thu, 1 Mar 2012 11:11:04 UTC

Hello, nquadr!

Quote:

Let X = someNumber, to help me solve this equation.

$24500(1+i)^3 - 10000(1+i)^{\frac{5}{3}} + 8000(1+i)^{\frac{13}{12}} \:=\: 30309$

$\text{Let } X \,=\,(1+i)^{12}$

$\text{Then we have: }\:24,\!500X^{36} - 10,\!000X^{20} + 8,\!000X^{13} - 30,\!309 \:=\:0$

. . Better?

Voxx · **Posted:** Thu, 1 Mar 2012 11:22:23 UTC

i suggested to pass 1+i to the trigonometric form and see if the simplification could be easier...

Shadow · **Posted:** Thu, 1 Mar 2012 16:04:11 UTC

Soroban wrote:

Hello, nquadr!

Quote:

Let X = someNumber, to help me solve this equation.

$24500(1+i)^3 - 10000(1+i)^{\frac{5}{3}} + 8000(1+i)^{\frac{13}{12}} \:=\: 30309$

$\text{Let } X \,=\,(1+i)^{12}$

$\text{Then we have: }\:24,\!500X^{36} - 10,\!000X^{20} + 8,\!000X^{13} - 30,\!309 \:=\:0$

. . Better?

I think you mean $$x=(1+i)^{1\over 12}$

Shadow · **Posted:** Thu, 1 Mar 2012 16:04:29 UTC

Voxx wrote:

i suggested to pass 1+i to the trigonometric form and see if the simplification could be easier...

It's not a complex number, if it were, this would be easy.

Denis · **Posted:** Thu, 1 Mar 2012 17:26:36 UTC

Soroban wrote:

$\text{Then we have: }\:24,\!500X^{36} - 10,\!000X^{20} + 8,\!000X^{13} - 30,\!309 \:=\:0$
. . Better?

No, since the OP is learning financial formulas :idea:

If someone is taking first driving lesson,
not the time to show him how to drag race down main street!

Shadow · **Posted:** Thu, 1 Mar 2012 19:23:59 UTC

Denis wrote:

Soroban wrote:

$\text{Then we have: }\:24,\!500X^{36} - 10,\!000X^{20} + 8,\!000X^{13} - 30,\!309 \:=\:0$
. . Better?

No, since the OP is learning financial formulas :idea:

If someone is taking first driving lesson,
not the time to show him how to drag race down main street!

I don't think that's the right metaphor for this. What Soroban is doing is not really that much higher of a skill level required, it's just slightly different methodology. So it's more like learning to drive a car and being asked to drive a motorcycle.

Soroban · **Posted:** Thu, 1 Mar 2012 19:36:33 UTC

Hello, Shadow!

Quote:

I think you mean $$x=(1+i)^{1\over 12}$

Of course I did . . . *slap head*

S.O.S. Mathematics CyberBoard

Let X = substitution

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