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Question Number 150965 by EDWIN88 last updated on 17/Aug/21

Answered by john_santu last updated on 17/Aug/21

$$\mathrm{R}=\frac{1}{2}\log {}_{10}\mathrm{x}=\frac{1}{2}{\mathrm{P}}^2$$$$\mathrm{S}=\frac{1}{2}\log {}_{10}\mathrm{y}=\frac{1}{2}{\mathrm{Q}}^2$$$$\iff \mathrm{P}+\mathrm{Q}+\mathrm{R}+\mathrm{S}=24$$$$\iff \mathrm{P}+\mathrm{Q}+\frac{1}{2}{\mathrm{P}}^2+\frac{1}{2}{\mathrm{Q}}^2=24$$$$\iff {\mathrm{P}}^2+2\mathrm{P}+{\mathrm{Q}}^2+2\mathrm{Q}=48$$$$\iff {(\mathrm{P}+1)}^2+{(\mathrm{Q}+1)}^2=50$$$$\mathrm{holds}\mathrm{for}(\mathrm{P},\mathrm{Q})=\{(0,6),(4,4),(6,0)\}$$$$\mathrm{when}(\mathrm{P},\mathrm{Q})=(4,4)\Rightarrow \mathrm{xy}={10}^{\mathrm{m}}$$$$\Rightarrow \mathrm{m}=\log {}_{10}\left(\mathrm{x}\right)+\log {}_{10}\left(\mathrm{y}\right)$$$$\Rightarrow \mathrm{m}=\sqrt{{\left(\log {}_{10}\left(\mathrm{x}\right)\right)}^2}+\sqrt{{\left(\log {}_{10}\left(\mathrm{y}\right)\right)}^2}$$$$\mathrm{m}={\mathrm{P}}^2+{\mathrm{Q}}^2=16+16=32$$$$\mathrm{when}(\mathrm{P},\mathrm{Q})=(6,0)$$$$\Rightarrow \mathrm{m}=6^2+0^2=36,\mathrm{this}\mathrm{is}\mathrm{same}$$$$\mathrm{to}\mathrm{when}(\mathrm{P},\mathrm{Q})=(0,6)$$$$\mathrm{therefore}\mathrm{m}=32\mathrm{or}\mathrm{m}=36$$

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