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x-3-3367-2-n-x-n-




Question Number 147566 by mathdanisur last updated on 21/Jul/21
x^3  + 3367 = 2^n   ⇒ x ; n = ?
$${x}^{\mathrm{3}} \:+\:\mathrm{3367}\:=\:\mathrm{2}^{\boldsymbol{{n}}} \:\:\Rightarrow\:{x}\:;\:{n}\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 21/Jul/21
x^3 +3367 = 2^n   Solution for x∈N.    x = ((2^n −3367))^(1/3)   n ≤ 11 ⇒ 2^n −3367 < 0    n = 12 :  x = ((2^(12) −3367))^(1/3)   x = ((2^n −3367))^(1/3)   x = ((729))^(1/3)  = (9^3 )^(1/3)  = 9  ⇒ x = 9, n = 12 is a solution
$${x}^{\mathrm{3}} +\mathrm{3367}\:=\:\mathrm{2}^{{n}} \\ $$$$\mathrm{Solution}\:\mathrm{for}\:{x}\in\mathbb{N}. \\ $$$$ \\ $$$${x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{2}^{{n}} −\mathrm{3367}} \\ $$$${n}\:\leqslant\:\mathrm{11}\:\Rightarrow\:\mathrm{2}^{{n}} −\mathrm{3367}\:<\:\mathrm{0} \\ $$$$ \\ $$$${n}\:=\:\mathrm{12}\:: \\ $$$${x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{12}} −\mathrm{3367}} \\ $$$${x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{2}^{{n}} −\mathrm{3367}} \\ $$$${x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{729}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{9}^{\mathrm{3}} }\:=\:\mathrm{9} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{9},\:{n}\:=\:\mathrm{12}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$
Commented by mathdanisur last updated on 22/Jul/21
thank you Sir
$${thank}\:{you}\:{Sir} \\ $$

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