# Thread: system

1. ## system

can i solve the system

x^n = 12
n^x = 4

i keep running into problems like n=1 and x= (LN(4))/0
(wew... don't let the perfecthacker see that)

or would the values just be undifind??

dan

2. Originally Posted by dan
can i solve the system

x^n = 12
n^x = 4

i keep running into problems like n=1 and x= (LN(4))/0
(wew... don't let the perfecthacker see that)

or would the values just be undifind??

dan
x=\sqrt[n]{12}

x=ln(4)/ln(n)

thus: ln(4)/ln(n)=\sqrt[n]{12}

then: ln(4)=ln(n)*\sqrt[n]{12}

therefore: ln(4)=ln(n^(12^1/n))

thus: 4=n^(12^1/n)

and then I'm stuck

3. Originally Posted by dan
can i solve the system

x^n = 12
n^x = 4

i keep running into problems like n=1 and x= (LN(4))/0
(wew... don't let the perfecthacker see that)

or would the values just be undifind??

dan
The best I can do for you is this:
From the second equation, if we take ln of both sides:
ln(n^x) = ln(4)

x ln(n) = ln(4)

x = ln(4)/ln(n)

Insert this into the first equation:

[ln(4)/ln(n)]^n = 12

This is, at least, an equation in one variable. Supposedly you can solve this numerically.

-Dan

4. I belive you can use Lambert function on,
(though I did not try it)

5. Originally Posted by ThePerfectHacker
I belive you can use Lambert function on,
(though I did not try it)
How does the last line follow from the one before?

6. Originally Posted by dan
can i solve the system

x^n = 12
n^x = 4

i keep running into problems like n=1 and x= (LN(4))/0
(wew... don't let the perfecthacker see that)

or would the values just be undifind??

dan
This can be done numerically, when one finds:

x~=8.1272946
n~=1.1859838

RonL

7. Originally Posted by CaptainBlack
How does the last line follow from the one before?
A mistake.

8. Originally Posted by CaptainBlack
This can be done numerically, when one finds:

x~=8.1272946
n~=1.1859838

RonL

hmmm thanks did you do that with a computer progam??

dan

9. Originally Posted by dan
hmmm thanks did you do that with a computer progam??

dan
Yes, but I was even lazier than normal, what I did was use the Excel solver
to minimise:

f(x,n)=(x^n-12)^2+(n^x-4)^2,

which since this is always >=0, has minima at the roots of the system.

RonL

10. Originally Posted by CaptainBlack
Yes, but I was even lazier than normal, what I did was use the Excel solver
to minimise:

f(x,n)=(x^n-12)^2+(n^x-4)^2,

which since this is always >=0, has minima at the roots of the system.

RonL
cool...those are probebly illagle for students....^_^