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.
Hello Arka
Welcome to Math Help Forum!By inspection, we can see that one solution is $\displaystyle \displaystyle x = y = 1$. But if $\displaystyle \displaystyle y \ne 1$, then we can take logs, knowing that $\displaystyle \displaystyle \log y \ne 0$:
$\displaystyle \displaystyle \left\{\begin {array}{l} x^y = y^x\\$\displaystyle \displaystyle \Rightarrow \left\{\begin{array}{l} y\log x = x\log y\\
x^2=y^3\end{array}$
2\log x = 3 \log y \end{array}$
$\displaystyle \displaystyle \Rightarrow \dfrac{\log x}{\log y}=\dfrac xy=\dfrac32 $, provided $\displaystyle \displaystyle \log y \ne 0$.
$\displaystyle \displaystyle \Rightarrow y = \tfrac23x$
$\displaystyle \displaystyle \Rightarrow y^3 = \tfrac{8}{27}x^3=x^2$
$\displaystyle \Rightarrow 8x^3 - 27x^2=0$
$\displaystyle \displaystyle \Rightarrow x = 0$ or $\displaystyle \displaystyle x = \tfrac{27}{8}$
$\displaystyle \displaystyle x=0$ clearly gives us problems with $\displaystyle \log x$. So we're left with the solution $\displaystyle \displaystyle x = \tfrac{27}{8},\;y = \tfrac94$.
I haven't looked at (ii). I think you could begin in a similar way.
Grandad
Hello ArkaYes, 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:
$\displaystyle y^3 = x^2$Substitute into the equation $\displaystyle x^y = y^x$:
$\displaystyle \Rightarrow y = x^{\frac23}$
$\displaystyle x^{(x^{\frac23})}=\big(x^{\frac23\big)^x}$If we now compare the exponents on each side of the equation, we get:
$\displaystyle =x^{(\frac23x)}$
$\displaystyle \Rightarrow x^{\frac23}=\frac23x$Cube both sides:
$\displaystyle \Rightarrow x^2 = \frac{8}{27}x^3$Then, as before, $\displaystyle x = ...$
Grandad