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Question Number 150797 by mathdanisur last updated on 15/Aug/21

x;y∈N ((18^((x^2 + y^2 )/2) )/9^(xy) ) = 2592 ⇒ find xy=?

$$\mathrm{x};\mathrm{y}\in\mathbb{N} \\ $$$$\frac{\mathrm{18}^{\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:+\:\boldsymbol{\mathrm{y}}^{\mathrm{2}} }{\mathrm{2}}} }{\mathrm{9}^{\boldsymbol{\mathrm{xy}}} }\:=\:\mathrm{2592}\:\:\Rightarrow\:\mathrm{find}\:\:\boldsymbol{\mathrm{xy}}=? \\ $$

Answered by nimnim last updated on 15/Aug/21

(((2×3^2 )^((x^2 +y^2 )/2) )/3^(2xy) )=32×81 ⇒2^((x^2 +y^2 )/2) ×3^(x^2 +y^2 −2xy) =2^5 ×3^4 ⇒((x^2 +y^2 )/2)=5 and x^2 +y^2 −2xy=4 ⇒x^2 +y^2 =10 and 2xy=10−4 ⇒ xy=3 ★

$$\frac{\left(\mathrm{2}×\mathrm{3}^{\mathrm{2}} \right)^{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}} }{\mathrm{3}^{\mathrm{2}{xy}} }=\mathrm{32}×\mathrm{81} \\ $$$$\Rightarrow\mathrm{2}^{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}} ×\mathrm{3}^{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}} =\mathrm{2}^{\mathrm{5}} ×\mathrm{3}^{\mathrm{4}} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{5}\:\:\:{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{10}\:\:{and}\:\:\:\mathrm{2}{xy}=\mathrm{10}−\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{xy}=\mathrm{3}\:\bigstar \\ $$

Commented by mathdanisur last updated on 15/Aug/21

Thank you Ser

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

Commented by puissant last updated on 15/Aug/21

very nice bro

$${very}\:{nice}\:{bro} \\ $$

Answered by MJS_new last updated on 15/Aug/21

((18^u )/9^v )=2592 (((2×3^2 )^u )/((3^2 )^v ))=2^5 ×3^4 2^u ×3^(2(u−v)) =2^5 ×3^4 u=5 2(u−v)=4 ⇒ v=3 ⇔ xy=3 [ ((x),(y) ) = ((1),(3) ) ∨ ((3),(1) )]

$$\frac{\mathrm{18}^{{u}} }{\mathrm{9}^{{v}} }=\mathrm{2592} \\ $$$$\frac{\left(\mathrm{2}×\mathrm{3}^{\mathrm{2}} \right)^{{u}} }{\left(\mathrm{3}^{\mathrm{2}} \right)^{{v}} }=\mathrm{2}^{\mathrm{5}} ×\mathrm{3}^{\mathrm{4}} \\ $$$$\mathrm{2}^{{u}} ×\mathrm{3}^{\mathrm{2}\left({u}−{v}\right)} =\mathrm{2}^{\mathrm{5}} ×\mathrm{3}^{\mathrm{4}} \\ $$$${u}=\mathrm{5} \\ $$$$\mathrm{2}\left({u}−{v}\right)=\mathrm{4}\:\Rightarrow\:{v}=\mathrm{3}\:\Leftrightarrow\:{xy}=\mathrm{3} \\ $$$$\left[\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:\vee\:\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix}\right] \\ $$

Commented by mathdanisur last updated on 15/Aug/21

nice Ser, thank you

$$\mathrm{nice}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you} \\ $$

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