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Question Number 8998 by tawakalitu last updated on 11/Nov/16

If 2^x = 3^y = 6^(−z) find the value of : (1/x) + (1/y) + (1/z)

$$\mathrm{If}\:\:\mathrm{2}^{\mathrm{x}} \:=\:\mathrm{3}^{\mathrm{y}} \:=\:\mathrm{6}^{−\mathrm{z}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\::\:\:\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{y}}\:+\:\frac{\mathrm{1}}{\mathrm{z}} \\ $$

Answered by Rasheed Soomro last updated on 12/Nov/16

If 2^x = 3^y = 6^(−z) find the value of : (1/x) + (1/y) + (1/z) −−−−−−−−−−−−−−−−−− 2^x = 3^y = 6^(−z) ⇒xlog2=ylog3=−zlog6 x=−z(((log6)/(log2))) , y=−z(((log6)/(log3))) x^(−1) ={−z(((log6)/(log2)))}^(−1) ⇒(1/x)=− ((log2)/(zlog6)) y^(−1) ={−z(((log6)/(log3)))}^(−1) ⇒(1/y)=− ((log3)/(zlog6)) (1/x) + (1/y) + (1/z)=− ((log2)/(zlog6))− ((log3)/(zlog6))+(1/z)=((−log2−log3+log6)/(zlog6)) =((log2^(−1) +log3^(−1) +log6)/(zlog6))=((log((1/2)×(1/3)×6))/(zlog6))=((log1)/(zlog6)) =(0/(zlog6))=0 (1/x) + (1/y) + (1/z)=0

$$\mathrm{If}\:\:\mathrm{2}^{\mathrm{x}} \:=\:\mathrm{3}^{\mathrm{y}} \:=\:\mathrm{6}^{−\mathrm{z}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\::\:\:\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{y}}\:+\:\frac{\mathrm{1}}{\mathrm{z}} \\ $$$$−−−−−−−−−−−−−−−−−− \\ $$$$\:\mathrm{2}^{\mathrm{x}} \:=\:\mathrm{3}^{\mathrm{y}} \:=\:\mathrm{6}^{−\mathrm{z}} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\mathrm{xlog2}=\mathrm{ylog3}=−\mathrm{zlog6} \\ $$$$\mathrm{x}=−\mathrm{z}\left(\frac{\mathrm{log6}}{\mathrm{log2}}\right)\:,\:\mathrm{y}=−\mathrm{z}\left(\frac{\mathrm{log6}}{\mathrm{log3}}\right) \\ $$$$\mathrm{x}^{−\mathrm{1}} =\left\{−\mathrm{z}\left(\frac{\mathrm{log6}}{\mathrm{log2}}\right)\right\}^{−\mathrm{1}} \Rightarrow\frac{\mathrm{1}}{\mathrm{x}}=−\:\frac{\mathrm{log2}}{\mathrm{zlog6}} \\ $$$$\mathrm{y}^{−\mathrm{1}} =\left\{−\mathrm{z}\left(\frac{\mathrm{log6}}{\mathrm{log3}}\right)\right\}^{−\mathrm{1}} \Rightarrow\frac{\mathrm{1}}{\mathrm{y}}=−\:\frac{\mathrm{log3}}{\mathrm{zlog6}} \\ $$$$\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{y}}\:+\:\frac{\mathrm{1}}{\mathrm{z}}=−\:\frac{\mathrm{log2}}{\mathrm{zlog6}}−\:\frac{\mathrm{log3}}{\mathrm{zlog6}}+\frac{\mathrm{1}}{\mathrm{z}}=\frac{−\mathrm{log2}−\mathrm{log3}+\mathrm{log6}}{\mathrm{zlog6}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{log2}^{−\mathrm{1}} +\mathrm{log3}^{−\mathrm{1}} +\mathrm{log6}}{\mathrm{zlog6}}=\frac{\mathrm{log}\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{6}\right)}{\mathrm{zlog6}}=\frac{\mathrm{log1}}{\mathrm{zlog6}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{0}}{\mathrm{zlog6}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{y}}\:+\:\frac{\mathrm{1}}{\mathrm{z}}=\mathrm{0} \\ $$

Commented by tawakalitu last updated on 11/Nov/16

Thank you sir.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Theara last updated on 19/Nov/16

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