Question Number 50802 by Pk1167156@gmail.com last updated on 20/Dec/18
Commented by Pk1167156@gmail.com last updated on 20/Dec/18
help soon please!
Commented by Abdo msup. last updated on 20/Dec/18
$$\mathrm{4}^{{x}^{\mathrm{2}} −{x}−\mathrm{6}} \:=\mathrm{7}\:\Leftrightarrow\left({x}^{\mathrm{2}} −{x}−\mathrm{6}\right){ln}\left(\mathrm{4}\right)={ln}\left(\mathrm{7}\right)\:\Leftrightarrow \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{6}\:=\frac{{ln}\left(\mathrm{7}\right)}{\left.{l}\left.{n}\right)\mathrm{4}\right)}\:\Leftrightarrow{x}^{\mathrm{2}} −{x}−\mathrm{6}\:={log}_{\mathrm{4}} \left(\mathrm{7}\right)\:\Rightarrow \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{6}−{log}_{\mathrm{4}} \left(\mathrm{7}\right)=\mathrm{0} \\ $$$$\Delta\:=\mathrm{1}−\mathrm{4}\left(−\mathrm{6}−{log}_{\mathrm{4}} \left(\mathrm{7}\right)\right)=\mathrm{25}\:+\mathrm{4}{log}_{\mathrm{4}} \left(\mathrm{7}\right)>\mathrm{0}\:\Rightarrow \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{25}+\mathrm{4}{log}_{\mathrm{4}} \left(\mathrm{7}\right)}}{\mathrm{2}}\:\:{and} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{25}+\mathrm{4}{log}_{\mathrm{4}} \left(\mathrm{7}\right)}}{\mathrm{2}} \\ $$
Answered by hassentimol last updated on 20/Dec/18
$$\mathrm{Answers}\:\mathrm{are}\:: \\ $$$$\mathbb{S}\:=\:\left\{{x}\:\:=\:\:\mathrm{0}.\mathrm{5}\:\:\underset{−} {+}\:\:\:\frac{\sqrt{\mathrm{25}+\mathrm{4}×\mathrm{log}_{\mathrm{4}} \left(\mathrm{7}\right)}\:}{\mathrm{2}}\right\} \\ $$
Commented by Pk1167156@gmail.com last updated on 20/Dec/18
solution please!
Answered by afachri last updated on 20/Dec/18
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}^{{x}^{\mathrm{2}} −{x}−\mathrm{6}} \:=\:\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{7}\right)\:\:\:\:\:\:=\:{x}^{\mathrm{2}} −{x}−\mathrm{6}\: \\ $$$${x}^{\mathrm{2}} −{x}−\left(\mathrm{6}+\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{7}\right)\right)\:=\:\mathrm{0} \\ $$$$\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{{x}}_{\mathrm{1},\mathrm{2}} \::\:\:\frac{−{b}\:\pm\:\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}\:}}{\mathrm{2}{a}} \\ $$$${x}_{\mathrm{1},\mathrm{2}\:} \:=\:\:\frac{\mathrm{1}\:\pm\:\sqrt{\mathrm{1}^{} \:+\:\:\mathrm{4}\left(\mathrm{6}+\mathrm{log}_{\mathrm{4}} \left(\mathrm{7}\right)\right)^{} }}{\mathrm{2}}\: \\ $$$${x}_{\mathrm{1},\mathrm{2}\:} \:=\:\:\frac{\mathrm{1}\:\pm\:\sqrt{\mathrm{25}\:+\:\mathrm{4log}\:_{\mathrm{4}} \left(\mathrm{7}\right)^{} }}{\mathrm{2}} \\ $$