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Question Number 145769 by Engr_Jidda last updated on 07/Jul/21
find x if 2^x +2^(3x) =16
$${find}\:{x}\:{if}\:\:\mathrm{2}^{{x}} +\mathrm{2}^{\mathrm{3}{x}} =\mathrm{16} \\ $$
Commented by 7770 last updated on 08/Jul/21
2^x +2^(3x) =16 let 2^x =m m+m^3 =16 m^3 +m−16=0 let f(x)=m^3 +m−16 f^′ (x)=3m^2 +1 let the initial guess m be 0 using nri m_1 =m_0 −((f(x))/(f^′ (x))) m=2.387686553... 2^x =m 2^x =2.387686553... x=((ln2.387686553...)/(ln2)) x≈1.255613
$$\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{2}^{\mathrm{3}\boldsymbol{{x}}} =\mathrm{16} \\ $$$$\:\boldsymbol{{let}}\:\mathrm{2}^{\boldsymbol{{x}}} =\boldsymbol{{m}} \\ $$$$\boldsymbol{{m}}+\boldsymbol{{m}}^{\mathrm{3}} =\mathrm{16} \\ $$$$\:\boldsymbol{{m}}^{\mathrm{3}} +\boldsymbol{{m}}−\mathrm{16}=\mathrm{0} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{m}}^{\mathrm{3}} +\boldsymbol{{m}}−\mathrm{16} \\ $$$$\:\boldsymbol{{f}}^{'} \left(\boldsymbol{{x}}\right)=\mathrm{3}\boldsymbol{{m}}^{\mathrm{2}} +\mathrm{1} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{the}}\:\boldsymbol{{initial}}\:\boldsymbol{{guess}}\:\boldsymbol{{m}}\:\boldsymbol{{be}}\:\mathrm{0} \\ $$$$\boldsymbol{{using}}\:\boldsymbol{{nri}}\:\: \\ $$$$\:\boldsymbol{{m}}_{\mathrm{1}} =\boldsymbol{{m}}_{\mathrm{0}} −\frac{\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{f}}^{'} \left(\boldsymbol{{x}}\right)} \\ $$$$\:\boldsymbol{{m}}=\mathrm{2}.\mathrm{387686553}… \\ $$$$\:\:\mathrm{2}^{\boldsymbol{{x}}} =\boldsymbol{{m}} \\ $$$$\:\mathrm{2}^{\boldsymbol{{x}}} =\mathrm{2}.\mathrm{387686553}… \\ $$$$\:\boldsymbol{{x}}=\frac{\boldsymbol{{ln}}\mathrm{2}.\mathrm{387686553}…}{\boldsymbol{{ln}}\mathrm{2}} \\ $$$$\:\boldsymbol{{x}}\approx\mathrm{1}.\mathrm{255613} \\ $$
Answered by mr W last updated on 07/Jul/21
let t=2^x t^3 +t−16=0 Δ=(1/(27))+64>0 (√Δ)=((√(5187))/9) ⇒t=((((√(5187))/9)+8))^(1/3) −((((√(5187))/9)−8))^(1/3) ⇒x=log_2 (((((√(5187))/9)+8))^(1/3) −((((√(5187))/9)−8))^(1/3) ) ≈1.255613
$${let}\:{t}=\mathrm{2}^{{x}} \\ $$$${t}^{\mathrm{3}} +{t}−\mathrm{16}=\mathrm{0} \\ $$$$\Delta=\frac{\mathrm{1}}{\mathrm{27}}+\mathrm{64}>\mathrm{0} \\ $$$$\sqrt{\Delta}=\frac{\sqrt{\mathrm{5187}}}{\mathrm{9}} \\ $$$$\Rightarrow{t}=\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{5187}}}{\mathrm{9}}+\mathrm{8}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{5187}}}{\mathrm{9}}−\mathrm{8}} \\ $$$$\Rightarrow{x}=\mathrm{log}_{\mathrm{2}} \:\left(\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{5187}}}{\mathrm{9}}+\mathrm{8}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{5187}}}{\mathrm{9}}−\mathrm{8}}\right) \\ $$$$\:\:\:\:\:\:\:\:\approx\mathrm{1}.\mathrm{255613} \\ $$
Commented by Engr_Jidda last updated on 08/Jul/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$

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