Exponential simultaneous equations

**Arka** · July 15th 2010, 10:21 PM

A little help with these would be of help!

i) x^y= y^x
x^2=y^3

Solve for x and y.

ii)a^x=x^y
a^y=x^x

Same here.

**Grandad** · July 16th 2010, 01:01 AM

Hello Arka

Welcome to Math Help Forum!

Originally Posted by Arka

A little help with these would be of help!

i) x^y= y^x
x^2=y^3

Solve for x and y.

ii)a^x=x^y
a^y=x^x

Same here.

By inspection, we can see that one solution is $\displaystyle x = y = 1$ . But if $\displaystyle y \ne 1$ , then we can take logs, knowing that $\displaystyle \log y \ne 0$ :

$\displaystyle \left\{\begin {array}{l} x^y = y^x\\<br /> x^2=y^3\end{array}$

$\displaystyle \Rightarrow \left\{\begin{array}{l} y\log x = x\log y\\<br /> 2\log x = 3 \log y \end{array}$

$\displaystyle \Rightarrow \dfrac{\log x}{\log y}=\dfrac xy=\dfrac32$ , provided $\displaystyle \log y \ne 0$ .

$\displaystyle \Rightarrow y = \tfrac23x$

$\displaystyle \Rightarrow y^3 = \tfrac{8}{27}x^3=x^2$

$\Rightarrow 8x^3 - 27x^2=0$

$\displaystyle \Rightarrow x = 0$ or $\displaystyle x = \tfrac{27}{8}$

$\displaystyle x=0$ clearly gives us problems with $\log x$ . So we're left with the solution $\displaystyle x = \tfrac{27}{8},\;y = \tfrac94$ .

I haven't looked at (ii). I think you could begin in a similar way.

Grandad

**Arka** · July 16th 2010, 01:25 AM

Cant we solve it without using logarithms? Coz in this sum , my prof said that we are not allowed to use log!

**Grandad** · July 16th 2010, 02:07 AM

Hello Arka

Originally Posted by Arka

Cant we solve it without using logarithms? Coz in this sum , my prof said that we are not allowed to use log!

Yes, you can; although it would have been helpful if you had said so at the beginning.

I prefer to use logs, because it avoids handling fractional exponents, but here goes: