Question Number 42019 by Tawa1 last updated on 16/Aug/18
$$\mathrm{find}\:\mathrm{x}:\:\:\:\:\:\mathrm{2}^{\mathrm{x}} \:+\:\mathrm{3}^{\mathrm{x}} \:=\:\mathrm{13} \\ $$
Commented by math khazana by abdo last updated on 17/Aug/18
$${let}\:{f}\left({x}\right)=\mathrm{2}^{{x}} \:+\mathrm{3}^{{x}} \:−\mathrm{13}\:\:\:\left({e}\right)\Leftrightarrow\:{f}\left({x}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)={e}^{{xln}\left(\mathrm{2}\right)} \:+{e}^{{xln}\mathrm{3}} −\mathrm{13}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:={ln}\left(\mathrm{2}\right)\:\mathrm{2}^{{x}} \:\:+{ln}\left(\mathrm{3}\right)\mathrm{3}^{{x}} \:>\mathrm{0}\:\:{so}\:{f}\:\:{is}\:{increasing} \\ $$$${on}\:{R}\:\:\:{we}\:{have}\:{lim}_{{x}\:\rightarrow−\infty} {f}\left({x}\right)=−\mathrm{13}\:\:{and} \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=+\infty\:\:\:{so}\:\exists\:{x}_{\mathrm{0}} \:\in{R}\:/{f}\left({x}_{\mathrm{0}} \right)=\mathrm{0} \\ $$$${we}\:{have}\:{f}\left(\mathrm{0}\right)\:=−\mathrm{13}\:<\mathrm{0}\:\:{and}\:\:{f}\left(\mathrm{3}\right)=\mathrm{2}^{\mathrm{3}} \:+\mathrm{3}^{\mathrm{3}} −\mathrm{13} \\ $$$$\left.=\mathrm{8}\:+\mathrm{27}−\mathrm{13}\:=\mathrm{22}>\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{0}} \in\right]\mathrm{0},\mathrm{3}\left[\:\:{we}\:{can}\right. \\ $$$${verify}\:{that}\:{x}_{\mathrm{0}} =\mathrm{2}\:\:{is}\:{exact}\:{solution}\:{to}\:{f}\left({x}\right)=\mathrm{0} \\ $$
Commented by Tawa1 last updated on 17/Aug/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by maxmathsup by imad last updated on 17/Aug/18
$${thanks}\:{sir} \\ $$
Answered by MJS last updated on 17/Aug/18
$${x}=\mathrm{2} \\ $$
Commented by Tawa1 last updated on 17/Aug/18
$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{but}\:\mathrm{any}\:\mathrm{workings}\:?? \\ $$
Commented by MJS last updated on 17/Aug/18
$$\mathrm{2}^{{x}} +\mathrm{3}^{{x}} =\mathrm{13} \\ $$$${x}=\frac{\mathrm{ln}\:{t}}{\mathrm{ln}\:\mathrm{2}} \\ $$$${t}^{\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}} +{t}=\mathrm{13} \\ $$$$\mathrm{since}\:\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}\:\mathrm{is}\:\mathrm{irrational}\:\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$
Commented by Tawa1 last updated on 17/Aug/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$