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Question Number 21470 by Joel577 last updated on 24/Sep/17

$$2^x=3^y=6^{-z}$$$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}$$$${(\frac{2017}{x}+\frac{2017}{y}+\frac{2017}{z})}^{2017}$$

Commented by Joel577 last updated on 24/Sep/17

$$2^x=3^y=6^{-z}=k(k\ne 0)$$$$2^x=k\to x={}^2\log k$$$$3^y=k\to y={}^3\log k$$$$6^{-z}=k\to -z={}^6\log k\to z={}^6\log k^{-1}$$$$$$$${(\frac{{}^2\log 2^{2017}}{{}^2\log k}+\frac{{}^3\log 3^{2017}}{{}^3\log k}+\frac{{}^6\log 6^{2017}}{{}^6\log k^{-1}})}^{2017}$$$$={({}^k\log 2^{2017}+{}^k\log 3^{2017}+{}^{k^{-1}}\log 6^{2017})}^{2017}$$$$={({}^k\log 2^{2017}+{}^k\log 3^{2017}-{}^k\log 6^{2017})}^{2017}$$$$={\left[{}^k\log \left(\frac{2^{2017}.3^{2017}}{6^{2017}}\right)\right]}^{2017}$$$$={\left({}^k\log 1\right)}^{2017}$$$$=0$$

Answered by $@ty@m last updated on 24/Sep/17

$$2^x=3^y=6^{-z}=k,say$$$$\Rightarrow k^{\frac{1}{x}}=2,k^{\frac{1}{y}}=3\mathrm{\&}{k}^{\frac{-1}{z}}=6$$$$\because 2\times 3=6$$$$\therefore k^{\frac{1}{x}}\times k^{\frac{1}{y}}=k^{\frac{-1}{z}}$$$$\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}$$$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$$$\frac{2017}{x}+\frac{2017}{y}+\frac{2017}{z}=0$$$$$$

Commented by Joel577 last updated on 26/Sep/17

$$thankyoufortheshortcut$$

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