# Solve the system

• Aug 19th 2009, 01:06 PM
dhiab
Solve the system
Solve the system :

$\left\{ \begin{array}{l}
x^y = y^x \\
a^x = b^y \\
\end{array} \right.$
• Aug 21st 2009, 04:19 AM
Kiwi_Dave
I hope that you like logarithms. Hopefully there may be an easier solution than mine!

From equation #2
$x=log_a(b^y)=ylog_ab$

Substituting for x in equation #1
$\therefore (ylog_ab)^y=y^{ylog_ab}$

$\therefore y^y(log_ab)^y=(y^y)^{log_ab}$

$\therefore (log_ab)^y=(y^y)^{(log_ab-1)}$

taking log of both sides to base $log_ab$

$\therefore log_{log_ab}((log_ab)^y)=y=log_{log_ab}(y^y)^{(log _ab-1)}=y(log_ab-1)log_{log_ab}(y)$

dividing through by y

$\therefore (log_ab-1)log_{log_ab}(y)=1$

$\therefore log_{log_ab}(y)=\frac 1{log_ab-1}$

$\therefore y=log_ab$ to the power of $(\frac 1{log_ab-1})$

Example; say a = e =2.718 and b = 17 then y=1.764875.

$x=log_a(b^y)=5.000267$ from equation #2

finally

$x^y=y^x=17.1252$
• Aug 22nd 2009, 09:14 AM
Wilmer
Quote:

Originally Posted by dhiab
Solve the system :
$\left\{ \begin{array}{l}
x^y = y^x \\
a^x = b^y \\
\end{array} \right.$

Why can't x = y = 2 and a = b = any number n ?
2^2 = 2^2
n^2 = n^2
• Aug 24th 2009, 01:46 AM
Kiwi_Dave
It is usual for letters from the first half of the alphabet to be considered as constants and letters from the last half of the alphabet to be variables. Therefore, if the question had been written more formally it might have said something like:

Solve the following system of equations for x and y, given the constants a and b.
• Aug 24th 2009, 05:32 AM
Wilmer
Quote:

Originally Posted by Kiwi_Dave
It is usual.....

That to me means no more than "it is usual for men to wear pants
and wowen to wear dresses".
"It is usual" is quite dangerous in mathematics...
In programming, using a to m as variables is not "syntax error".

Ain't arguing, mate, just my opinion.