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Question Number 139861 by mathdanisur last updated on 01/May/21

((2z+1))^(1/3) + ((8z+4))^(1/5) = (√(16))

$$\sqrt[{\mathrm{3}}]{\mathrm{2}\boldsymbol{{z}}+\mathrm{1}}\:+\:\sqrt[{\mathrm{5}}]{\mathrm{8}\boldsymbol{{z}}+\mathrm{4}}\:=\:\sqrt{\mathrm{16}} \\ $$

Answered by mindispower last updated on 01/May/21

z=(7/2),solution over R f(z)=((2z+1))^(1/3) +((8z+4))^(1/5) is increase function lim_(z→−∞) f(z)=−∞ lim_(z→+∞) f(z)=+∞,f((7/2))=4 S={(7/2)}

$${z}=\frac{\mathrm{7}}{\mathrm{2}},{solution} \\ $$$${over}\:\mathbb{R} \\ $$$${f}\left({z}\right)=\sqrt[{\mathrm{3}}]{\mathrm{2}{z}+\mathrm{1}}+\sqrt[{\mathrm{5}}]{\mathrm{8}{z}+\mathrm{4}}\:{is}\:{increase}\:{function} \\ $$$$\underset{{z}\rightarrow−\infty} {\mathrm{lim}}{f}\left({z}\right)=−\infty \\ $$$$\underset{{z}\rightarrow+\infty} {\mathrm{lim}}{f}\left({z}\right)=+\infty,{f}\left(\frac{\mathrm{7}}{\mathrm{2}}\right)=\mathrm{4} \\ $$$${S}=\left\{\frac{\mathrm{7}}{\mathrm{2}}\right\} \\ $$$$ \\ $$$$ \\ $$

Commented by mathdanisur last updated on 02/May/21

thanks sir

$${thanks}\:{sir} \\ $$

Answered by MJS_new last updated on 02/May/21

t=2z+1 ⇔ z=((t−1)/2) t^(1/3) +4^(1/5) t^(1/5) =4 t=u^(15) u^5 +4^(1/5) u^3 −4=0 u^5 +2^(2/5) u^3 −4=0 looks a lot like u=2^v 2^(5v) +2^(2/5) 2^(3v) =4 2^(5v) +2^(3v+2/5) =4 v=1/5 ⇒ u=2^(1/5) ⇒ t=8 ⇒ z=(7/2)

$${t}=\mathrm{2}{z}+\mathrm{1}\:\Leftrightarrow\:{z}=\frac{{t}−\mathrm{1}}{\mathrm{2}} \\ $$$${t}^{\mathrm{1}/\mathrm{3}} +\mathrm{4}^{\mathrm{1}/\mathrm{5}} {t}^{\mathrm{1}/\mathrm{5}} =\mathrm{4} \\ $$$${t}={u}^{\mathrm{15}} \\ $$$${u}^{\mathrm{5}} +\mathrm{4}^{\mathrm{1}/\mathrm{5}} {u}^{\mathrm{3}} −\mathrm{4}=\mathrm{0} \\ $$$${u}^{\mathrm{5}} +\mathrm{2}^{\mathrm{2}/\mathrm{5}} {u}^{\mathrm{3}} −\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{looks}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{like}\:{u}=\mathrm{2}^{{v}} \\ $$$$\mathrm{2}^{\mathrm{5}{v}} +\mathrm{2}^{\mathrm{2}/\mathrm{5}} \mathrm{2}^{\mathrm{3}{v}} =\mathrm{4} \\ $$$$\mathrm{2}^{\mathrm{5}{v}} +\mathrm{2}^{\mathrm{3}{v}+\mathrm{2}/\mathrm{5}} =\mathrm{4} \\ $$$${v}=\mathrm{1}/\mathrm{5}\:\Rightarrow\:{u}=\mathrm{2}^{\mathrm{1}/\mathrm{5}} \:\Rightarrow\:{t}=\mathrm{8}\:\Rightarrow\:{z}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$

Commented by mathdanisur last updated on 02/May/21

thankyou sir

$${thankyou}\:{sir} \\ $$

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