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Question-50802




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
4^(x^2 −x−6)  =7 ⇔(x^2 −x−6)ln(4)=ln(7) ⇔  x^2 −x−6 =((ln(7))/(ln)4))) ⇔x^2 −x−6 =log_4 (7) ⇒  x^2 −x−6−log_4 (7)=0  Δ =1−4(−6−log_4 (7))=25 +4log_4 (7)>0 ⇒  x_1 =((1+(√(25+4log_4 (7))))/2)  and  x_2 =((1−(√(25+4log_4 (7))))/2)
$$\mathrm{4}^{{x}^{\mathrm{2}} −{x}−\mathrm{6}} \:=\mathrm{7}\:\Leftrightarrow\left({x}^{\mathrm{2}} −{x}−\mathrm{6}\right){ln}\left(\mathrm{4}\right)={ln}\left(\mathrm{7}\right)\:\Leftrightarrow \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{6}\:=\frac{{ln}\left(\mathrm{7}\right)}{\left.{l}\left.{n}\right)\mathrm{4}\right)}\:\Leftrightarrow{x}^{\mathrm{2}} −{x}−\mathrm{6}\:={log}_{\mathrm{4}} \left(\mathrm{7}\right)\:\Rightarrow \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{6}−{log}_{\mathrm{4}} \left(\mathrm{7}\right)=\mathrm{0} \\ $$$$\Delta\:=\mathrm{1}−\mathrm{4}\left(−\mathrm{6}−{log}_{\mathrm{4}} \left(\mathrm{7}\right)\right)=\mathrm{25}\:+\mathrm{4}{log}_{\mathrm{4}} \left(\mathrm{7}\right)>\mathrm{0}\:\Rightarrow \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{25}+\mathrm{4}{log}_{\mathrm{4}} \left(\mathrm{7}\right)}}{\mathrm{2}}\:\:{and} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{25}+\mathrm{4}{log}_{\mathrm{4}} \left(\mathrm{7}\right)}}{\mathrm{2}} \\ $$
Answered by hassentimol last updated on 20/Dec/18
Answers are :  S = {x  =  0.5  +_−    (((√(25+4×log_4 (7))) )/2)}
$$\mathrm{Answers}\:\mathrm{are}\:: \\ $$$$\mathbb{S}\:=\:\left\{{x}\:\:=\:\:\mathrm{0}.\mathrm{5}\:\:\underset{−} {+}\:\:\:\frac{\sqrt{\mathrm{25}+\mathrm{4}×\mathrm{log}_{\mathrm{4}} \left(\mathrm{7}\right)}\:}{\mathrm{2}}\right\} \\ $$
Commented by Pk1167156@gmail.com last updated on 20/Dec/18
solution please!
Answered by afachri last updated on 20/Dec/18
                                 4^(x^2 −x−6)  = 7                              log _4 (7)      = x^2 −x−6   x^2 −x−(6+log _4 (7)) = 0  to find x_(1,2)  :  ((−b ± (√(b^2 −4ac )))/(2a))  x_(1,2 )  =  ((1 ± (√(1^  +  4(6+log_4 (7))^ )))/2)   x_(1,2 )  =  ((1 ± (√(25 + 4log _4 (7)^ )))/2)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}^{{x}^{\mathrm{2}} −{x}−\mathrm{6}} \:=\:\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{7}\right)\:\:\:\:\:\:=\:{x}^{\mathrm{2}} −{x}−\mathrm{6}\: \\ $$$${x}^{\mathrm{2}} −{x}−\left(\mathrm{6}+\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{7}\right)\right)\:=\:\mathrm{0} \\ $$$$\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{{x}}_{\mathrm{1},\mathrm{2}} \::\:\:\frac{−{b}\:\pm\:\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}\:}}{\mathrm{2}{a}} \\ $$$${x}_{\mathrm{1},\mathrm{2}\:} \:=\:\:\frac{\mathrm{1}\:\pm\:\sqrt{\mathrm{1}^{} \:+\:\:\mathrm{4}\left(\mathrm{6}+\mathrm{log}_{\mathrm{4}} \left(\mathrm{7}\right)\right)^{} }}{\mathrm{2}}\: \\ $$$${x}_{\mathrm{1},\mathrm{2}\:} \:=\:\:\frac{\mathrm{1}\:\pm\:\sqrt{\mathrm{25}\:+\:\mathrm{4log}\:_{\mathrm{4}} \left(\mathrm{7}\right)^{} }}{\mathrm{2}} \\ $$

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