Zombie Apocalypse Logs

Using the formula

$\log_{a}{N}=\frac{\log_{b}{N}}{\log{_{b}{a}}},$

we can rewrite our system of equation in a following way:

$\begin{cases} \log_{2}{x}+\frac{1}{2}\log_{2}{y}+\frac{1}{2}\log_{2}{z}=2 \\ \log_{3}{y}+\frac{1}{2}\log_{3}{z}+\frac{1}{2}\log_{3}{x}=2 \\ \log_{4}{z}+\frac{1}{2}\log_{4}{x}+\frac{1}{2}\log_{4}{y}=2 \end{cases}$

Simplifying $^1$ , we obtain

$\begin{cases} x\sqrt{yz}=4 \\ y\sqrt{zx}=9 \\z\sqrt{xy}=16 \end{cases}$ $(a)$

Multiplying all of the equations out, we get

$(xyz)^2=4\cdot 9\cdot 16$

from which

$xyz=24.$ $(b)$

(from the given problem we take the principal root of the equation in order to make sense for the original $x,y,$ and $z$ arguments of the logs).

Squaring each equation in system $(a)$ and dividing them by $(b)$ , we obtain

$\boxed{x=\frac{2}{3},y=\frac{27}{8},z=\frac{32}{3}}.$

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1 - simplification took place in a following way:

$\begin{cases} \log_{2}{x}+\frac{1}{2}\log_{2}{y}+\frac{1}{2}\log_{2}{z}=2 \\ \log_{3}{y}+\frac{1}{2}\log_{3}{z}+\frac{1}{2}\log_{3}{x}=2 \\ \log_{4}{z}+\frac{1}{2}\log_{4}{x}+\frac{1}{2}\log_{4}{y}=2 \end{cases} \Rightarrow \begin{cases} \log _{ 2 }{ x\sqrt { yz } } =2 \\ \log _{ 3 }{ y\sqrt { zx } } =2 \\ \log _{ 4 }{ z\sqrt { xy } } =2 \end{cases}\Rightarrow \begin{cases} 2^{\log_{2}{x\sqrt{yz}}}=2^2 \\ 3^{\log_{3}{y\sqrt{zx}}}=3^2 \\ 4^{\log_{4}{z\sqrt{xy}}}=4^2 \end{cases}$

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s

log 2⁡x+log 4⁡y+log 4⁡z = 2 log 2⁡x+log 4⁡y+log 4⁡z = (log x/ log 2) + (log y/ log4) + (log z/log4) = (log x/ log 2) + (log y/ 2log2) + (log z/2log2) = (1/2log2)(logx^2 + log y + log z) .

    (1/2log2)(logx^2 + log y + log z)  = 2.
    logx^2yz = log 16.
    x^2yz= 16  - (1)...

Similarly xy^2z=81 - (2)... xyz^2=256 - (3)... multiplying (1), (2), (3) 〖(xyz)〗^4= 〖(2 4 3)〗^4. xyz = 24. eqn(1) becomes x(xyz) = 16. eqn (2) Becomes y(xyz) = 81. eqn(3) Becomes z(xyz) = 256. Substitute xyz in above equations x = 16/24 = 2/3. y = 81/24 = 27/8. z = 256/24 = 32/3. therefore, 3x+8y+3z = 3(2/3)+8(27/8)+3(32/3) = 3 + 27 + 32 = 61.