System

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Hello

I need help, solving the system.

$$\left ( 1+y \right )^{x}= 100$
$$\left ( y^{4}- 2y^{2}+1 \right )^{x-1}= \frac{\left ( y-1 \right )^{2x}}{\left ( y+1 \right )^{2}}$

Attempted:

$$\left[(y^2-1)^{x-1}\right]^2=\left[\cfrac{(y-1)^x}{(y+1)}\right]^2 \quad \quad |\;\sqrt$

$$\frac{(y^2-1)^{x}}{(y^2-1)}=\frac{(y-1)^x}{(y+1)}$

I'm not really sure about what should I be doing with these kind of problems ,
I don't know if I started it right.. but I'd be delighted, if I could solve this one (with your help).. Thank you

Shadow · **Posted:** Mon, 12 Mar 2012 20:53:29 UTC

kreshnik wrote:

Hello

I need help, solving the system.

$$\left ( 1+y \right )^{x}= 100$
$$\left ( y^{4}- 2y^{2}+1 \right )^{x-1}= \frac{\left ( y-1 \right )^{2x}}{\left ( y+1 \right )^{2}}$

Attempted:

$$\left[(y^2-1)^{x-1}\right]^2=\left[\cfrac{(y-1)^x}{(y+1)}\right]^2 \quad \quad |\;\sqrt$

$$\frac{(y^2-1)^{x}}{(y^2-1)}=\frac{(y-1)^x}{(y+1)}$

I'm not really sure about what should I be doing with these kind of problems ,
I don't know if I started it right.. but I'd be delighted, if I could solve this one (with your help).. Thank you

??

You have $(y-1)^{x-1}(y+1)}^{x-1}={(y-1)^{2(x-1)+2}\over (y+1)^2}$

and (y+1)^x=100

If y=1 you get one thing, if not, cancel in the second given:

$100={(y-1)^{x+1}\over (y+1)^3}$ , and that's about as far as I can see it going. Your attempt somehow adds a square into the mix that isn't there in the statement of the problem. Either you made a typo in the problem statement and this is actually quite simple, or you did not and that's as far as you can go.

Denis · **Posted:** Tue, 13 Mar 2012 06:16:37 UTC

kreshnik wrote:

Attempted:
$$\left[(y^2-1)^{x-1}\right]^2=\left[\cfrac{(y-1)^x}{(y+1)}\right]^2 \quad \quad |\;\sqrt$
$$\frac{(y^2-1)^{x}}{(y^2-1)}=\frac{(y-1)^x}{(y+1)}$

Good work; continue....since y^2-1 = (y-1)(y+1):

[(y+1)(y-1)]^x / [(y+1)(y-1)] = (y-1)^x / (y+1) ; since (y+1)^x = 100:

100(y-1)^x / [(y+1)(y-1)] = (y-1)^x / (y+1) ; cross multiply:

100(y+1)(y-1)^x = (y+1)(y-1)(y-1)^x ; do cancellations:

100 = y - 1
y = 101

(y+1)^x = 100
102^x = 100
x = LOG(100) / LOG(102) = .99571832568.....

Shadow · **Posted:** Tue, 13 Mar 2012 06:22:59 UTC

Denis wrote:

kreshnik wrote:

Attempted:
$$\left[(y^2-1)^{x-1}\right]^2=\left[\cfrac{(y-1)^x}{(y+1)}\right]^2 \quad \quad |\;\sqrt$
$$\frac{(y^2-1)^{x}}{(y^2-1)}=\frac{(y-1)^x}{(y+1)}$

Good work; continue....since y^2-1 = (y-1)(y+1):

[(y+1)(y-1)]^x / [(y+1)(y-1)] = (y-1)^x / (y+1) ; since (y+1)^x = 100:

100(y-1)^x / [(y+1)(y-1)] = (y-1)^x / (y+1) ; cross multiply:

100(y+1)(y-1)^x = (y+1)(y-1)(y-1)^x ; do cancellations:

100 = y - 1
y = 101

(y+1)^x = 100
102^x = 100
x = LOG(100) / LOG(102) = .99571832568.....

And what about the magical appearance of a square that was not there before?

Denis · **Posted:** Tue, 13 Mar 2012 06:35:26 UTC

Shadow wrote:

And what about the magical appearance of a square that was not there before?

Where's that? Can't see it! All I see is y^4 - 2y^2 + 1 = (y^2 - 1)^2, which is ok.
My solution:
y = 101
x = .99571832568.....
checks out if substituted in 2nd equation.

Shadow · **Posted:** Tue, 13 Mar 2012 06:48:21 UTC

Denis wrote:

Shadow wrote:

And what about the magical appearance of a square that was not there before?

Where's that? Can't see it! All I see is y^4 - 2y^2 + 1 = (y^2 - 1)^2, which is ok.
My solution:
y = 101
x = .99571832568.....
checks out if substituted in 2nd equation.

XD, yeah not quite sure why I thought it wasn't there.

Denis · **Posted:** Tue, 13 Mar 2012 06:52:55 UTC

T's'ok...1st mistake this year!

Denis · **Posted:** Tue, 13 Mar 2012 15:18:56 UTC

kreshnik wrote:

$$\left ( 1+y \right )^{x}= 100$
$$\left ( y^{4}- 2y^{2}+1 \right )^{x-1}= \frac{\left ( y-1 \right )^{2x}}{\left ( y+1 \right )^{2}}$

Kreshnik, I find it often saves lots of time (and writing!) to go a bit like this:

Let a = y+1 and b = y-1 ; then a^x = 100 [1]

Since y^4 - 2y^2 + 1 = [(y+1)(y-1)]^2
then left side = (a^2 b^2)^(x-1) = [a^x b^x / (ab)]^2 [2]

right side = b^(2x) / a^2 = (b^x / a)^2 [3]

[2][3]: a^x b^x / (ab) = b^x / a
a^x b^x / b = b^x
[1] 100b^x / b = b^x
100b^x = (b)b^x
b = 100
y - 1 = 100
y = 101

Something for you to think about...

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Thank you everyone.

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