If $x^2yz^3 = 7^3$ and $xy^2 = 7^9$, what are the values of $xyz$?

Question

My math teacher gave our class a challenge question today on exponents, but none of us have any idea how to solve it. We've tried linear systems, but it never quite worked out. Any ideas?

Answer 1

The quickest way is probably $$(x^2yz^3)(xy^2)=(7^3)(7^9)\Leftrightarrow x^3y^3z^3=7^{12}\Leftrightarrow(xyz)^3=(7^4)^3\Rightarrow xyz=7^4.$$

Let's try another (more general) strategy: Since $a^b=e^{b\operatorname{ln}(a)},$ we can write the given equations as $$x^2yz^3 =e^{2\operatorname{ln}(x)}e^{\operatorname{ln}(y)}e^{3\operatorname{ln}(z)} =e^{2\operatorname{ln}(x)+\operatorname{ln}(y)+3\operatorname{ln}(z)} =7^3 =e^{3\operatorname{ln}(7)}$$

and

$$xy^2 =e^{\operatorname{ln}(x)}e^{2\operatorname{ln}(y)} =e^{\operatorname{ln}(x)+2\operatorname{ln}(y)} =7^9 =e^{9\operatorname{ln}(7)}.$$

That is, setting $a_1:=\operatorname{ln}(x),a_2:=\operatorname{ln}(y)$ and $a_3:=\operatorname{ln}(z),$ we get the system of linear equations

$$2a_1+a_2+3a_3=3\operatorname{ln}(7)$$ $$a_1+2a_2=9\operatorname{ln}(7)$$

or in matrix notation

$$ \underbrace{ \begin{pmatrix} 2 & 1 & 3\\ 1 & 2 & 0 \end{pmatrix}}_{=:A} \underbrace{ \begin{pmatrix} a_1\\a_2\\a_3 \end{pmatrix}}_{=: b} = \underbrace{ \begin{pmatrix} 3\operatorname{ln}(7)\\ 9\operatorname{ln}(7)\\ \end{pmatrix}}_{=:c}. $$ For the solution of this equation, we can use the Moore-Penrose inverse of $A$, which leads to $$b=A^T(A\cdot A^T)^{-1}c = \begin{pmatrix} \frac{1}{9} & \frac{1}{9}\\ -\frac{1}{18} & \frac{4}{9}\\ \frac{5}{18} & -\frac{2}{9} \end{pmatrix} \begin{pmatrix} 3\operatorname{ln}(7)\\ 9\operatorname{ln}(7)\\ \end{pmatrix} = \begin{pmatrix} \frac{4}{3}\operatorname{ln}(7)\\ \frac{23}{6}\operatorname{ln}(7)\\ -\frac{7}{6}\operatorname{ln}(7)\\ \end{pmatrix}. $$ This means $a_1=\operatorname{ln}(x)=\frac{4}{3}\operatorname{ln}(7),$ i.e. $x=e^{a_1}=7^{\frac{4}{3}}.$ Similarly we get $y=7^{\frac{23}{6}}$ and $z=7^{-\frac{7}{6}}.$ We can now compute $$xyz=7^{\frac{4}{3}}7^{\frac{23}{6}}7^{-\frac{7}{6}}=7^{\frac{24}{6}}=7^4.$$

Edit: Note, that there are other solutions to the matrix-equation by taking different pseudo-inverses of $A$. Thus, the values for $x,y$ and $z$ are not unique (but the result of $xyz$ is).

Answer 2

$$x^2yx^3\cdot xy^2=x^3y^3z^3=(xyz)^3=7^{12}$$ hence $$xyz=\sqrt[3]{7^{12}}=7^4=2401$$

Answer 3

As suggested in one of the comments, taking the (natural) logarithms, we obtain a system of $2$ linear equations in variables $\ln(x)$, $\ln(y)$ and $\ln(z)$. Its augmented matrix is the following

$$\left[\begin{array}{ccc|c} 2 & 1 & 3 & 3 \ln (7)\\ 1 & 2 & 0 & 9 \ln(7)\end{array}\right]$$

Using Gaussian elimination, we obtain the following row-reduced augmented matrix

$$\left[\begin{array}{ccc|c} 1 & 0 & 2 & - \ln(7)\\ 0 & 1 & -1 & 5 \ln(7)\end{array}\right]$$

Let $t := \ln(z)$ parameterize the solution set (a curve) in the positive octant. Hence,

$$\begin{array}{rl} \ln (x) &= - 2 t - \ln (7)\\ \ln (y) &= t + 5 \ln (7)\end{array}$$

and, thus,

$$\begin{array}{rl} x &= \frac 17 \, e^{-2t}\\ y &= 7^5 e^{t}\\ z &= e^t\end{array}$$

Lastly,

$$\boxed{ \,\\ \,\quad x y z = 7^4 \quad \\}$$