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2-x-3-y-4-2-y-3-x-6-prove-that-xy-z-z-1-6-z-2-




Question Number 173871 by mathlove last updated on 20/Jul/22
2^x ∙3^y =4  2^y ∙3^x =6       prove that  xy=z(z+1)  6^z =2
$$\mathrm{2}^{{x}} \centerdot\mathrm{3}^{{y}} =\mathrm{4} \\ $$$$\mathrm{2}^{{y}} \centerdot\mathrm{3}^{{x}} =\mathrm{6}\:\:\:\:\:\:\:{prove}\:{that}\:\:{xy}={z}\left({z}+\mathrm{1}\right) \\ $$$$\mathrm{6}^{{z}} =\mathrm{2} \\ $$
Commented by dragan91 last updated on 20/Jul/22
((2^x 3^y )/(2^y 3^x ))=(2/3)  ((2/3))^(x−y) =(2/3)⇒x−y=1⇒x=y+1  2^(y+1) 3^y =4  2^y 3^(y+1) =6  6^y =2⇒y=log_6 2⇒x=log_6 2+1  xy=log_6 2(log_6 2+1)  z=log_6 2  xy=z(z+1) proved
$$\frac{\mathrm{2}^{\mathrm{x}} \mathrm{3}^{\mathrm{y}} }{\mathrm{2}^{\mathrm{y}} \mathrm{3}^{\mathrm{x}} }=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}−\mathrm{y}} =\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow\mathrm{x}−\mathrm{y}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{y}+\mathrm{1} \\ $$$$\mathrm{2}^{\mathrm{y}+\mathrm{1}} \mathrm{3}^{\mathrm{y}} =\mathrm{4} \\ $$$$\mathrm{2}^{\mathrm{y}} \mathrm{3}^{\mathrm{y}+\mathrm{1}} =\mathrm{6} \\ $$$$\mathrm{6}^{\mathrm{y}} =\mathrm{2}\Rightarrow\mathrm{y}=\mathrm{log}_{\mathrm{6}} \mathrm{2}\Rightarrow\mathrm{x}=\mathrm{log}_{\mathrm{6}} \mathrm{2}+\mathrm{1} \\ $$$$\mathrm{xy}=\mathrm{log}_{\mathrm{6}} \mathrm{2}\left(\mathrm{log}_{\mathrm{6}} \mathrm{2}+\mathrm{1}\right) \\ $$$$\mathrm{z}=\mathrm{log}_{\mathrm{6}} \mathrm{2} \\ $$$$\mathrm{xy}=\mathrm{z}\left(\mathrm{z}+\mathrm{1}\right)\:\mathrm{proved} \\ $$$$ \\ $$

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