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Question Number 129951 by Algoritm last updated on 21/Jan/21

Commented by MJS_new last updated on 21/Jan/21

2^(x/3) −2^x =2 t=2^(x/3) t−t^3 =2 t≈−1.52∨t≈.761±.858i 2^(2x/3) +2^(4x/3) +4^x =5 t=2^(x/3) t^6 +t^4 +t^2 =5 t≈±1.13∨t≈±(.648±1.25i) 2^x −8^x =10 (?) t=2^(x/3) t^3 −t^9 =10 t≈−1.32∨t≈−.960±.841∨t≈−.249±1.25∨t≈.661±1.14 ⇒ no solution at all

$$\mathrm{2}^{{x}/\mathrm{3}} −\mathrm{2}^{{x}} =\mathrm{2} \\ $$$${t}=\mathrm{2}^{{x}/\mathrm{3}} \\ $$$${t}−{t}^{\mathrm{3}} =\mathrm{2} \\ $$$${t}\approx−\mathrm{1}.\mathrm{52}\vee{t}\approx.\mathrm{761}\pm.\mathrm{858i} \\ $$$$ \\ $$$$\mathrm{2}^{\mathrm{2}{x}/\mathrm{3}} +\mathrm{2}^{\mathrm{4}{x}/\mathrm{3}} +\mathrm{4}^{{x}} =\mathrm{5} \\ $$$${t}=\mathrm{2}^{{x}/\mathrm{3}} \\ $$$${t}^{\mathrm{6}} +{t}^{\mathrm{4}} +{t}^{\mathrm{2}} =\mathrm{5} \\ $$$${t}\approx\pm\mathrm{1}.\mathrm{13}\vee{t}\approx\pm\left(.\mathrm{648}\pm\mathrm{1}.\mathrm{25i}\right) \\ $$$$ \\ $$$$\mathrm{2}^{{x}} −\mathrm{8}^{{x}} =\mathrm{10}\:\left(?\right) \\ $$$${t}=\mathrm{2}^{{x}/\mathrm{3}} \\ $$$${t}^{\mathrm{3}} −{t}^{\mathrm{9}} =\mathrm{10} \\ $$$${t}\approx−\mathrm{1}.\mathrm{32}\vee{t}\approx−.\mathrm{960}\pm.\mathrm{841}\vee{t}\approx−.\mathrm{249}\pm\mathrm{1}.\mathrm{25}\vee{t}\approx.\mathrm{661}\pm\mathrm{1}.\mathrm{14} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{at}\:\mathrm{all} \\ $$

Answered by SEKRET last updated on 21/Jan/21

10

$$\mathrm{10} \\ $$

Answered by SEKRET last updated on 21/Jan/21

2^(x/2) =a a−a^3 =2 a^2 +a^4 +a^6 =5 (a−a^3 )(a^2 +a^4 +a^6 )=2∙5 a^3 −a^9 =10

$$\:\mathrm{2}^{\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}} =\boldsymbol{\mathrm{a}} \\ $$$$\:\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{a}}^{\mathrm{3}} =\mathrm{2} \\ $$$$\:\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{6}} =\mathrm{5} \\ $$$$\:\:\left(\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{a}}^{\mathrm{3}} \right)\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{4}} +\boldsymbol{\mathrm{a}}^{\mathrm{6}} \right)=\mathrm{2}\centerdot\mathrm{5} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\boldsymbol{\mathrm{a}}^{\mathrm{9}} =\mathrm{10} \\ $$

Commented by MJS_new last updated on 21/Jan/21

great new method! { ((2^y =9)),((3^y =4)) :} 2^y ×3^y =9×4 6^y =36 ⇒ y=2

$$\mathrm{great}\:\mathrm{new}\:\mathrm{method}! \\ $$$$\begin{cases}{\mathrm{2}^{{y}} =\mathrm{9}}\\{\mathrm{3}^{{y}} =\mathrm{4}}\end{cases} \\ $$$$\mathrm{2}^{{y}} ×\mathrm{3}^{{y}} =\mathrm{9}×\mathrm{4} \\ $$$$\mathrm{6}^{{y}} =\mathrm{36} \\ $$$$\Rightarrow\:{y}=\mathrm{2} \\ $$

Answered by bemath last updated on 21/Jan/21

(2^(x/3) −2^x )(2^((2x)/3) +2^((4x)/3) +2^(2x) )=10 2^x +2^((5x)/3) +2^((7x)/3) −2^((5x)/3) −2^((7x)/3) −2^(3x) =10 2^x −2^(3x) = 10 2^x −8^x = 10

$$\left(\mathrm{2}^{\frac{\mathrm{x}}{\mathrm{3}}} −\mathrm{2}^{\mathrm{x}} \right)\left(\mathrm{2}^{\frac{\mathrm{2x}}{\mathrm{3}}} +\mathrm{2}^{\frac{\mathrm{4x}}{\mathrm{3}}} +\mathrm{2}^{\mathrm{2x}} \right)=\mathrm{10} \\ $$$$\:\mathrm{2}^{\mathrm{x}} +\mathrm{2}^{\frac{\mathrm{5x}}{\mathrm{3}}} +\mathrm{2}^{\frac{\mathrm{7x}}{\mathrm{3}}} −\mathrm{2}^{\frac{\mathrm{5x}}{\mathrm{3}}} −\mathrm{2}^{\frac{\mathrm{7x}}{\mathrm{3}}} −\mathrm{2}^{\mathrm{3x}} \:=\mathrm{10} \\ $$$$\:\mathrm{2}^{\mathrm{x}} −\mathrm{2}^{\mathrm{3x}} =\:\mathrm{10} \\ $$$$\:\mathrm{2}^{\mathrm{x}} −\mathrm{8}^{\mathrm{x}} \:=\:\mathrm{10} \\ $$

Commented by MJS_new last updated on 21/Jan/21

people, this is wrong! there′s no x solving both equations as I showed. multiplying two equations is not allowed in case the variables are not the same { ((x=5)),((x=3)) :} ⇒^? x^2 =15

$$\mathrm{people},\:\mathrm{this}\:\mathrm{is}\:\mathrm{wrong}!\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:{x}\:\mathrm{solving} \\ $$$$\mathrm{both}\:\mathrm{equations}\:\mathrm{as}\:\mathrm{I}\:\mathrm{showed}.\:\mathrm{multiplying} \\ $$$$\mathrm{two}\:\mathrm{equations}\:\mathrm{is}\:\mathrm{not}\:\mathrm{allowed}\:\mathrm{in}\:\mathrm{case}\:\mathrm{the} \\ $$$$\mathrm{variables}\:\mathrm{are}\:\mathrm{not}\:\mathrm{the}\:\mathrm{same} \\ $$$$\begin{cases}{{x}=\mathrm{5}}\\{{x}=\mathrm{3}}\end{cases} \\ $$$$\overset{?} {\Rightarrow}\:{x}^{\mathrm{2}} =\mathrm{15} \\ $$

Commented by liberty last updated on 21/Jan/21

do you meant prof this question is false ?

$$\mathrm{do}\:\mathrm{you}\:\mathrm{meant}\:\mathrm{prof}\:\mathrm{this}\:\mathrm{question} \\ $$$$\mathrm{is}\:\mathrm{false}\:? \\ $$

Commented by MJS_new last updated on 21/Jan/21

obviously

$$\mathrm{obviously} \\ $$

Commented by liberty last updated on 21/Jan/21

yes. i agree

$$\mathrm{yes}.\:\mathrm{i}\:\mathrm{agree} \\ $$

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