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Question Number 80780 by naka3546 last updated on 06/Feb/20

$$2\cdot m^x+3\cdot n^y=18$$$$min\{m^x\cdot n^y\}=?$$

Commented by naka3546 last updated on 06/Feb/20

$$m,n,x,y\in \mathbb{N}$$

Answered by key of knowledge last updated on 06/Feb/20

$$\mathrm{n},\mathrm{y},\mathrm{m},\mathrm{x}\in \mathrm{N}\Rightarrow {\mathrm{m}}^{\mathrm{x}}=\mathrm{a},{\mathrm{n}}^{\mathrm{y}}=\mathrm{b}\in \mathrm{N}$$$$2\mathrm{a}+3\mathrm{b}=18\Rightarrow \{\begin{array}{l}\mathrm{a}={\mathrm{m}}^{\mathrm{x}}=3=3^1\mathrm{b}={\mathrm{n}}^{\mathrm{y}}=4=2^2\Rightarrow \min ={\mathrm{m}}^{\mathrm{x}}\\ \mathrm{a}={\mathrm{m}}^{\mathrm{x}}=6=6^1\mathrm{b}={\mathrm{n}}^{\mathrm{y}}=2=2^1\Rightarrow \min ={\mathrm{n}}^{\mathrm{y}}\end{array}$$$$$$

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