Complicated System of Equations involving Logarithms

Question

I am trying to solve this system of equations. I know the answer but I am struggling with the working. I need to find $x$, $y$, and $z$ in terms of $a$, $b$, and $c$. The system of equations is shown below: \begin{align} n&=\log a+p+\log z\\ n&=\log b+\log y+q\\ n&=\log x+\log c+r\\ n&=\log a+\log b+\log x\\ n&=p+\log y+\log c\\ n&=\log z+q+r\\ n&=\log a+\log y+r\\ n&=\log x+\log y+\log z \end{align} I know: \begin{align} x&=\sqrt{a}\cdot\sqrt{b}\\ y&=\sqrt{a}\cdot\sqrt{b}\\ z&=\sqrt{a}\cdot\sqrt{b} \end{align}

I would appreciate some help with working.

P.S. Thanks to @Haris Gusic for editing the LaTeX for me,

JUST TO CLARIFY: I know I have provided the answer I am looking for a method to get to the answer as I simply used a calculator to find it but I would like to know how to do via full working.

Answer 1

It is simpler to get rid of non-logarithmic terms by mapping $n\mapsto\log n$ and similarly for $p,q,r$ obtaining: $$\begin{align} n&=apz\tag1\\ n&=byq\tag2\\ n&=xcr\tag3\\ n&=abx\tag4\\ n&=pyc\tag5\\ n&=zqr\tag6\\ n&=ayr\tag7\\ n&=xyz\tag8\\ \end{align} $$ where all numbers are assumed to be positive.

(2)+(5)+(7): $$bq=pc=ar=\frac ny.\tag9$$

(9)+(1):$$n=az\frac n{yc}\implies z=\frac ca y\tag{10}$$

(9)+(3):$$n=xc\frac n{ya}\implies x=\frac ac y\tag{11}$$

(10)+(11)+(8): $$n=y^3\tag{12}$$

Substituting the values of $x$ and $n$ from (11) and (12) into (4) one obtains: $$ y^2=a^2\frac bc\tag{13} $$ and finally combining (10), (11) and (13): $$ x=a^2b^\frac12c^{-\frac32},\quad y=ab^\frac12c^{-\frac12},\quad z=b^\frac12c^{\frac12}. $$

---

Note that replacing the equation (4) with a similar $n=abz$ one would obtain $x=y=z=a^\frac12b^{\frac12}$ in agreement with the claim of OP. This suggests that there is a misprint in at least one of the original equations.

Answer 2

\begin{align} n&=\log a+p+\log z \tag{1}\label{1} \\ n&=\log b+\log y+q \tag{2}\label{2} \\ n&=\log x+\log c+r \tag{3}\label{3} \\ n&=\log a+\log b+\log x \tag{4}\label{4} \\ n&=p+\log y+\log c \tag{5}\label{5} \\ n&=\log z+q+r \tag{6}\label{6} \\ n&=\log a+\log y+r \tag{7}\label{7} \\ n&=\log x+\log y+\log z \tag{8}\label{8} . \end{align}

If we rename all terms of the form $\log v$ as a new parameter $v'$, we'll get a system of eight linear equations with ten parameters, $a',b',c',p,q,r,x',y',z',n$.

We can reorder all these equations
to move $a',b'$ to the right (assuming $c'$ as unknown for the moment) and get the system $\mathbf{S_8}$ in a matrix form

\begin{align} \mathbf{A_8}\cdot\mathbf{X_8}&=\mathbf{B_8} , \end{align}

where \begin{align} \mathbf{A_8}&= \begin{bmatrix} 0& 0& -1& 1& -1& 0& 0& 0 \\ 0& -1& 0& 1& 0& -1& 0& 0 \\ -1& 0& 0& 1& 0& 0& -1& -1 \\ -1& 0& 0& 1& 0& 0& 0& 0 \\ 0& -1& 0& 1& -1& 0& 0& -1 \\ 0& 0& -1& 1& 0& -1& -1& 0 \\ 0& -1& 0& 1& 0& 0& -1& 0 \\ -1& -1& -1& 1& 0& 0& 0& 0 \end{bmatrix} ,\\ \mathbf{X_8}&= \begin{bmatrix} x'& y'& z'& n& p& q& r& c' \end{bmatrix}^{\mathsf{T}} ,\\ \mathbf{B_8}&= \begin{bmatrix} a'& b'& 0& a'+b'& 0& 0& a'& 0 \end{bmatrix}^{\mathsf{T}} . \end{align}

We can see that \eqref{3} is a linear combination of the other:

\begin{align} \eqref{3}&= \eqref{5}+\eqref{6}+ \eqref{4}-\eqref{1}-\eqref{2} , \end{align} hence, \eqref{3} can be ignored.

Also, subtracting \eqref{3} from \eqref{4}, we can express $r$ in terms of $a',b',c'$: \begin{align} r&=a'+b'-c' , \end{align}

so we can construct a simpler linear system $\mathbf{S_7}$:

\begin{align} \mathbf{A_7}&= \begin{bmatrix} 0& 0& -1& 1& -1& 0& 0 \\ 0& -1& 0& 1& 0& -1& 0 \\ -1& 0& 0& 1& 0& 0& 0 \\ 0& -1& 0& 1& -1& 0& -1 \\ 0& 0& -1& 1& 0& -1& 1 \\ 0& -1& 0& 1& 0& 0& 1 \\ -1& -1& -1& 1& 0& 0& 0 \end{bmatrix} ,\\ \mathbf{X_7}&= \begin{bmatrix} x'& y'& z'& n& p& q& c' \end{bmatrix}^{\mathsf{T}} ,\\ \mathbf{B_7}&= \begin{bmatrix} a'& b'& a'+b'& 0& a'+b'& 2a'+b'& 0 \end{bmatrix}^{\mathsf{T}} . \end{align}

From rows 6 and 2 of the new system, we conclude that \begin{align} q&=2a'-c' . \end{align}

Also, from rows 5 and 1, we conclude that \begin{align} p&=b'+q-c'=b'+2a'-2c' , \end{align}

and we get a system $\mathbf{S_4}$: \begin{align} \begin{bmatrix} 0& 0& 1& -1 \\ 0& 1& 0& -1 \\ 1& 0& 0& -1 \\ 1& 1& 1& -1 \end{bmatrix} \cdot \begin{bmatrix} x' \\ y' \\ z' \\ n \end{bmatrix} &= \begin{bmatrix} -3 a'-b'+2 c' \\ -2 a'-b'+c' \\ -a'-b' \\ 0 \end{bmatrix} , \end{align} which is easy to solve and get

\begin{align} n &= 3 a'+\tfrac32 b'-\tfrac32 c' ,\\ x'&=2 a'+\tfrac12 b'-\tfrac32 c' ,\\ y'&=a'+\tfrac12 b'-\tfrac12 c' ,\\ z'&=\tfrac12 b'+\tfrac12 c' . \end{align}

And hence

\begin{align} x&=\frac{a^2\sqrt{b}}{\sqrt{c^3}} ,\\ y&=\frac{a\sqrt{b}}{\sqrt c} ,\\ z&=\sqrt{bc} ,\\ p&=\log\frac{a^2b}{c^2} ,\\ q&=\log\frac{a^2}c ,\\ r&=\log\frac{ab}c . \end{align}

As expected, substitution of these values into equations \eqref{1}-\eqref{8} gives \begin{align} n&=3\log a+\tfrac32\log b-\tfrac32\log c \end{align} for all the eight equations.

Answer 3

Consider the following problem:

Problem (1): Given $n \in \mathbb{R}$, determine all matrices $A \in \mathbb{R}^{3\times 3}$ such that the following conditions hold:

1. the sum of elements of every row of $A$ is $n$,
2. the sum of elements of every column of $A$ is $n$,
3. the sum of elements of every diagonal of $A$ is $n$.

Solution: $A \in \mathbb{R}^{3\times 3}$ satisfies conditions 1., 2., 3. if and only if it is expressible as $$ A = \begin{bmatrix} \frac{n}{3} + \epsilon_{0,0} & \frac{n}{3} - 2\epsilon_{0,0} - \epsilon_{1,0} & \frac{n}{3} + \epsilon_{0,0} + \epsilon_{1,0} \\ \frac{n}{3} + \epsilon_{1,0} & \frac{n}{3} & \frac{n}{3} - \epsilon_{1,0}\\ \frac{n}{3} - \epsilon_{0,0} - \epsilon_{1,0} & \frac{n}{3} +2 \epsilon_{0,0} + \epsilon_{1,0} & \frac{n}{3} - \epsilon_{0,0}\\ \end{bmatrix}, $$ for $\epsilon_{0,0}$ and $\epsilon_{1,0} \in \mathbb{R}$.

My solution to the original problem uses the above result on the matrix $$ A_0 = \begin{bmatrix} \log a & p & \log z\\ \log b & \log y & q\\ \log x & \log c & r\\ \end{bmatrix}. $$

We can immediately express $x, y, z$ in terms of $a, b, n$:

$$ \log x - \frac{n}{3} = - \left(\log a - \frac{n}{3}\right) - \left(\log b - \frac{n}{3}\right) \iff \log x = n - \log ab \iff \boxed{x = \frac{e^{n}}{ab}}, $$

$$ \log y = \frac{n}{3} \iff \boxed{y = e^{\frac{n}{3}}}, $$

$$ \log z - \frac{n}{3} = - \left(\log x - \frac{n}{3}\right) \iff \log z = - \log x \iff \boxed{z = \frac{ab}{e^{\frac{n}{3}}}}. $$ Then by expressing also $c$ in terms of $a,b,n$ we have $$ \log c - \frac{n}{3} = 2 \left(\log a - \frac{n}{3}\right) + \left(\log b - \frac{n}{3}\right) \iff \\ \iff \log c = -2\frac{n}{3} + \log a^2 b, $$ that we can invert to express $n$ in terms of $a,b,c$: $$ 2\frac{n}{3} = \log \frac{a^2 b}{c}\iff\\ \iff \boxed{\frac{n}{3} = \log \left(\frac{a b^{\frac{1}{2}}}{c^{\frac{1}{2}}}\right)}. $$

Finally, by substituting $n$ in the previous equations we get:

• $x =\displaystyle \frac{\left(a b^{\frac{1}{2}}c^{-\frac{1}{2}}\right)^3}{ab} = a^2 b^{\frac{1}{2}} c^{-\frac{3}{2}} = \boxed{\displaystyle\frac{a^2 \sqrt{b}}{\sqrt{c^3}}}$,
• $y = \left(a b^{\frac{1}{2}}c^{-\frac{1}{2}}\right) = a b^{\frac{1}{2}}c^{-\frac{1}{2}} = \boxed{\displaystyle\frac{a \sqrt{b}}{\sqrt{c}}} $,
• $z = \frac{ab}{\left(a b^{\frac{1}{2}}c^{-\frac{1}{2}}\right)} = b^{\frac{1}{2}}c^{\frac{1}{2}} =\boxed{\displaystyle \sqrt{b}\sqrt{c}}$.

---

APPENDIX

Proof of solution to Problem (1)

Given the matrix expressed in terms of $\epsilon_{0,0}$ and $\epsilon_{1,0}$, it is straightforward to see that it satisfies the conditions 1., 2., 3.. Now let's demonstrate the vice-versa.

We must determine $9$ variables with $8$ equations, so we expect that the space of solutions has dimension at least $1$. Note, however, that the sum of the elements of the third column is equal to the sum of the elements of the three rows minus the sum of the elements of the two columns, so at least one of the eight equations is redundant; hence the space of solutions has dimension at least $2$. Denote with $a_{i,j}$ the element of the $i$-th row and $j$-th column of $A$. Note that $a_{i,j} = n/3\ \forall i, \forall j$ is a solution. Therefore without losing generality we can fix $n = 3$ and express every element as $a_{i,j} = 1 + \epsilon_{i,j} $. Now let's fix $\epsilon_{0,0}$ and demonstrate the following

Lemma: In the above context $\epsilon_{2,2} = -\epsilon_{0,0}$.

Proof: We have that: $$ a_{1,0}+a_{2,0} = a_{1,1}+a_{2,2} = a_{0,1}+a_{0,2} = n - a_{0,0}, $$ that in terms of epsilons becomes: $$ 2+\epsilon_{1,0}+\epsilon_{2,0} = 2+\epsilon_{1,1}+\epsilon_{2,2} = 2+\epsilon_{0,1}+\epsilon_{0,2} = 2 - \epsilon_{0,0}\iff\\ \iff \epsilon_{1,0}+\epsilon_{2,0} = \epsilon_{1,1}+\epsilon_{2,2} = \epsilon_{0,1}+\epsilon_{0,2} = - \epsilon_{0,0}. $$ Then we can determine $a_{1,2}+a_{2,1}$ by subtracting from the sum of all elements, the couples found at previous step: $$ a_{1,2}+a_{2,1} = 9 - (a_{1,0}+a_{2,0}) - (a_{1,1}+a_{2,2}) - (a_{0,1}+a_{0,2}) - a_{0,0}\iff\\ \iff 2+\epsilon_{1,2}+\epsilon_{2,1} = 9 - (2 - \epsilon_{0,0}) - (2 - \epsilon_{0,0}) - (2 - \epsilon_{0,0}) - (1 + \epsilon_{0,0})\iff\\ \iff \epsilon_{1,2}+\epsilon_{2,1} = 2\epsilon_{0,0}. $$ With analog considerations we can deduce that $\epsilon_{0,1} +\epsilon_{1,0} = 2 \epsilon_{2,2}$. By subtracting the five elements on the diagonals from the sum of all elements we have: $$ 9-3 - (1+\epsilon_{0,0})- (1+\epsilon_{2,2}) = (2+\epsilon_{0,1} +\epsilon_{1,0}) + (2+\epsilon_{1,2}+\epsilon_{2,1})\iff\\ \iff- \epsilon_{0,0}- \epsilon_{2,2} = 2\epsilon_{0,0} + 2\epsilon_{2,2}\iff\\ \iff \epsilon_{2,2} = - \epsilon_{0,0}.$$

Immediately follows the

Corollary: In the above context $\epsilon_{1,1} = 0$.

Proof:

$$ a_{0,0} + a_{1,1}+a_{2,2} = 3 \iff \epsilon_{0,0} + \epsilon_{1,1}+\epsilon_{2,2} = 0 \iff\epsilon_{1,1} = -\epsilon_{0,0} - \epsilon_{2,2} =0. $$

Since we expect the space of solutions to have dimension at least $2$, we can try to fix two elements of the matrix and deduce the others.

$$ \begin{bmatrix} \cdot & \cdot & \cdot \\ \cdot & 1 & \cdot\\ \cdot & \cdot & \cdot\\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 + \epsilon_{0,0} & \cdot & \cdot \\ \cdot & 1 & \cdot\\ \cdot & \cdot & \cdot\\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 + \epsilon_{0,0} & \cdot & \cdot \\ \cdot & 1 & \cdot\\ \cdot & \cdot & 1 - \epsilon_{0,0}\\ \end{bmatrix} \rightarrow\\ \rightarrow \begin{bmatrix} 1 + \epsilon_{0,0} & \cdot & \cdot \\ 1 + \epsilon_{1,0} & 1 & \cdot\\ \cdot & \cdot & 1 - \epsilon_{0,0}\\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 + \epsilon_{0,0} & \cdot & \cdot \\ 1 + \epsilon_{1,0} & 1 & 1 - \epsilon_{1,0}\\ 1 - \epsilon_{0,0} -\epsilon_{1,0} & \cdot & 1 - \epsilon_{0,0}\\ \end{bmatrix} \rightarrow\\ \rightarrow \begin{bmatrix} 1 + \epsilon_{0,0} & 1 -2\epsilon_{0,0} -\epsilon_{1,0} & 1 + \epsilon_{0,0} +\epsilon_{1,0} \\ 1 + \epsilon_{1,0} & 1 & 1 - \epsilon_{1,0}\\ 1 - \epsilon_{0,0} -\epsilon_{1,0} & 1 +2\epsilon_{0,0} +\epsilon_{1,0} & 1 - \epsilon_{0,0}\\ \end{bmatrix}. $$ As we can see from the previous deductions, $2$ elements are enough for deducing all other elements.