# Thread: Solve the System of equations

1. ## Solve the System of equations

Been trying to work these two problems for the last couple hours without to much success anyone able to show me how they solved them. Would really help me get this assignment completed thanks.

ln(x^2*z^3/y) = 1
ln(x^4*y/z) = 2
ln(y^3 * z^5/x^2) = 3

and

e^x+e^y+e^z = 3
3e^x+2e^y-5e^z = 0
e^x-2e^y+3e^z = 2

Been trying to work these two problems for the last couple hours without to much success anyone able to show me how they solved them. Would really help me get this assignment completed thanks.

ln(x^2*z^3/y) = 1
ln(x^4*y/z) = 2
ln(y^3 * z^5/x^2) = 3

and

e^x+e^y+e^z = 3
3e^x+2e^y-5e^z = 0
e^x-2e^y+3e^z = 2
for the first one, raise everything to the power of e:
e^(ln(x^2*z^3/y)) = e
e^(ln(x^4*y/z)) = e^2
e^(ln(y^3 * z^5/x^2)) = e^3

this reduces down to:
x^2*z^3/y = e
x^4*y/z = e^2
y^3 * z^5/x^2 = e^3

From here, solve like a normal system of equations system.

The same goes for the second set, except take the natural log of everything. This works because:
e^(ln(x)) = x
ln(e^x) = x

3. Any chance you can finish up the last part. The first part looks good Thanks alot.

4. I was able to solve question #2 by a different method. I used substitution. I'm still trying to figure out #1 if anyone else could provide some input it would be much appreciated.

Let A=e^x, B=e^y & C=e^z

A+B+C=3 (1)
3A+2B-5C=0 (2)
A-2B+3C=2 (3)

Equation (2) dded to Eq. (3)
3A+2B-5C=0 (2)
A-2B+3C=2
Gives us:
4A-2C = 2

Eq. (1) Mult. by -2 added to Eq. (2)
-2A-2B-2C=-6 (1)
3A+2B-5C=0
Gives us:
A-7C=-6

So we are left with the system:
A+B+C=3
4A-2C=2
A-7C=-6

From this point it all comes together. The Answer: A=1, B=1 & C=1

$\displaystyle \begin{array}{ccc}\ln\left(\dfrac{x^2z^3}{y}\right ) &= &1\\ [5mm] \ln\left(\dfrac{x^4y}{z}\right) &=& 2 \\ [5mm] \ln\left(\dfrac{y^3z^5}{x^2}\right) &= &3 \end{array}$

$\displaystyle \ln\left(\frac{x^2z^3}{y}\right) \:=\:1\quad\Rightarrow\quad \ln(x^2) - \ln(y) + \ln(z^3) \:=\:1 \quad\Rightarrow$ . $\displaystyle 2\ln(x) - \ln(y) + 3\ln(z) \:=\:1$

$\displaystyle \ln\left(\frac{x^4y}{z}\right) \:=\:2\quad\Rightarrow\quad\ln(x^4) + \ln(y) - \ln(z) \:=\:2 \quad\Rightarrow$ . $\displaystyle 4\ln(x) + \ln(y) - \ln(z) \:=\:2$

$\displaystyle \ln\left(\frac{y^3z^5}{x^2}\right)\:=\:3\quad\Righ tarrow\quad \ln(y^3) + \ln(z^5) - \ln(x^2) \:=\:3\quad\Rightarrow$ . $\displaystyle -2\ln(x) + 3\ln(y) + 5\ln(5) \:=\:3$

Let: .$\displaystyle X = \ln(x),\;Y = \ln(y),\;Z = \ln(z)$

. . and we have: . $\displaystyle \begin{array}{ccc}2 - Y + 3Z &=& 1 \\ 4X + Y - Z &=& 2 \\ \text{-}2X + 3Y + 5Z &=& 3 \end{array}$

Solve the system: .$\displaystyle X = \frac{15}{38},\;Y = \frac{14}{19},\;Z = \frac{6}{19}$

Back-substitute: . $\displaystyle \begin{array}{ccccccc} \ln(x) \:=\:\dfrac{15}{38} & \Longrightarrow & {\color{blue}x \:=\: e^{\frac{15}{38}}} \\ \\ [-3mm] \ln(y) \:=\: \dfrac{14}{19} & \Longrightarrow & {\color{blue}y \:=\: e^{\frac{14}{19}}} \\ \\ [-3mm] \ln(z) \:=\: \dfrac{6}{19} & \Longrightarrow & {\color{blue}z \:=\: e^{\frac{6}{19}}} \end{array}$

6. Whoa. That's what I needed. Thanks alot Soroban. I was on the right track but I couldn't finish it. Thanks again.