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Question Number 24938 by kosarrr last updated on 29/Nov/17

3^y =2  2^x =3  4^(xy+1) ×9^(xy) =  24^(xy+1) =

$$\mathrm{3}^{{y}} =\mathrm{2} \\ $$$$\mathrm{2}^{{x}} =\mathrm{3} \\ $$$$\mathrm{4}^{{xy}+\mathrm{1}} ×\mathrm{9}^{{xy}} = \\ $$$$\mathrm{24}^{{xy}+\mathrm{1}} = \\ $$

Commented by prakash jain last updated on 29/Nov/17

4^(xy+1) =4×4^(xy) =4×2^(2xy) =4/((2^x )^y )^2   =4×(3^y )^2 =4×2^2 =16  =3^(2y) ×4=(3^y )^2 ×4=2^2 ×4=16  9^(xy) =3^(2xy) =((3^y )^x )^2 =(2^x )^2 =3^2 =9  4^(xy+1) ×9^(xy) =16×9=144

$$\mathrm{4}^{{xy}+\mathrm{1}} =\mathrm{4}×\mathrm{4}^{{xy}} =\mathrm{4}×\mathrm{2}^{\mathrm{2}{xy}} =\mathrm{4}/\left(\left(\mathrm{2}^{{x}} \right)^{{y}} \right)^{\mathrm{2}} \\ $$$$=\mathrm{4}×\left(\mathrm{3}^{{y}} \right)^{\mathrm{2}} =\mathrm{4}×\mathrm{2}^{\mathrm{2}} =\mathrm{16} \\ $$$$=\mathrm{3}^{\mathrm{2}{y}} ×\mathrm{4}=\left(\mathrm{3}^{{y}} \right)^{\mathrm{2}} ×\mathrm{4}=\mathrm{2}^{\mathrm{2}} ×\mathrm{4}=\mathrm{16} \\ $$$$\mathrm{9}^{{xy}} =\mathrm{3}^{\mathrm{2}{xy}} =\left(\left(\mathrm{3}^{{y}} \right)^{{x}} \right)^{\mathrm{2}} =\left(\mathrm{2}^{{x}} \right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} =\mathrm{9} \\ $$$$\mathrm{4}^{{xy}+\mathrm{1}} ×\mathrm{9}^{{xy}} =\mathrm{16}×\mathrm{9}=\mathrm{144} \\ $$

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