# Thread: another system of equations

1. ## another system of equations

solve for x,y

xy = e
x^(ln(y))=1/e^2

2. Hello, jmedsy!

A fascinating problem . . .

Solve for $x,y$

. . . $xy \:=\: e\;\;\;\;[1]$
. . $x^{\ln y} \:=\:\frac{1}{e^2}\;\;[2]$

Take logs of equation [1]: . $\log(xy) \:=\:\ln(e) \quad\Rightarrow\quad \ln(x) + \ln(y) \:=\:1]$

Take logs of equation [2]: . $\ln\left(x^{\ln y}\right) \:=\:\ln\left(\frac{1}{e^2}\right) \quad\Rightarrow\quad \ln(x)\cdot\ln(y) \:=\:-2$

We have a system of equations: . $\begin{array}{cccc} \ln(x) + \ln(y) &=& 1 \\ \ln(x)\cdot\ln(y) &=&-2 \end{array}$

Substitute: . $\begin{array}{ccc}u &=& \ln(x) \\ v &=& \ln(y) \end{array} \quad\Rightarrow\quad \begin{array}{cccc}u + v &=& 1 & [3] \\ u\cdot v &=& -2 & [4] \end{array}$

From [3], we have: . $v \:=\:1-u\;\;[5]$

Substitute into [4]: . $u(1-u) \:=\:-2 \quad\Rightarrow\quad u^2 - u - 2 \:=\:0$

Factor: . $(u+1)(u-2) \:=\:0 \quad\Rightarrow\quad u \:=\:\text{-}1,\:2$

. . $u = \text{-}1\!:\;\text{Substitute into [5]: }\:v \:=\:1-(\text{-}1) \quad\Rightarrow\quad v \,=\,2$

. . $u = 2\!:\;\text{Substitute into [5]: }\:v \:=\:1-2 \quad\Rightarrow\quad v \,=\,\text{-}1$

We have: . $(u,v) \:=\:(\text{-}1,2),\:(2,\text{-}1)$

Back-substitute:

. . $(u,v) \,=\,(\text{-}1,2) \quad\Rightarrow\quad \begin{array}{cccc}\ln(x) \;=\;\text{-}1 & \Rightarrow & x \;=\;\frac{1}{e} \\ \ln(y) \;=\; 2 & \Rightarrow & y \;=\; e^2 \end{array}$

. . $(u,v) \,=\,(2,\text{-}1) \quad\Rightarrow\quad \begin{array}{cccc}\ln(x) \;=\;2 & \Rightarrow & x \;=\;e^2 \\ \ln(y) \;=\;-1 & \Rightarrow & y \;=\;\frac{1}{e} \end{array}$

Two solutions: . $\left(\frac{1}{e^2},\;e^2\right),\;\left(e^2,\;\fr ac{1}{e}\right)$

3. finally came up with the same thing, thanks