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Question-200594




Question Number 200594 by cherokeesay last updated on 20/Nov/23
Answered by Frix last updated on 21/Nov/23
Obviously x=3 is a solution.  Then you must approximate... I found  x≈.211793616  x≈−2.80610974
$$\mathrm{Obviously}\:{x}=\mathrm{3}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}. \\ $$$$\mathrm{Then}\:\mathrm{you}\:\mathrm{must}\:\mathrm{approximate}…\:\mathrm{I}\:\mathrm{found} \\ $$$${x}\approx.\mathrm{211793616} \\ $$$${x}\approx−\mathrm{2}.\mathrm{80610974} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Nov/23
3^(x−1) .5^((3(x−1))/x) =3^(x^2 −7) .5^(x^2 −7)   3^(x^2 −7−x+1) .5^(x^2 −7−((3(x−1))/x)) =1  3^(x^2 −x−6) .5^((x^3 −10x+3)/x) =1  3^(x^2 −x−6) =5^(−((x^3 −10x+3)/x))   •  x^2 −x−6=0 ∧ −((x^3 −10x+3)/x)=0     (x−3)(x+2)=0 ∧ x^3 −10x+3=0     x=3,−2 ∧ (x−3)(x^2 +3x−1)=0   x=3,−2 ∧ x=3,((−3±(√(13)))/2)  ⇒x=3  •(x^2 −x−6)log3=(−((x^3 −10x+3)/x))log5      −((x^3 −10x+3)/(x(x^2 −x−6)))=((log3 )/(log5))=log_5 3  (log_5 3+1)x^3 −(log_5 3)x^2 −3(2log_5 3+5)x+3=0     ...
$$\mathrm{3}^{{x}−\mathrm{1}} .\mathrm{5}^{\frac{\mathrm{3}\left({x}−\mathrm{1}\right)}{{x}}} =\mathrm{3}^{{x}^{\mathrm{2}} −\mathrm{7}} .\mathrm{5}^{{x}^{\mathrm{2}} −\mathrm{7}} \\ $$$$\mathrm{3}^{{x}^{\mathrm{2}} −\mathrm{7}−{x}+\mathrm{1}} .\mathrm{5}^{{x}^{\mathrm{2}} −\mathrm{7}−\frac{\mathrm{3}\left({x}−\mathrm{1}\right)}{{x}}} =\mathrm{1} \\ $$$$\mathrm{3}^{{x}^{\mathrm{2}} −{x}−\mathrm{6}} .\mathrm{5}^{\frac{{x}^{\mathrm{3}} −\mathrm{10}{x}+\mathrm{3}}{{x}}} =\mathrm{1} \\ $$$$\mathrm{3}^{{x}^{\mathrm{2}} −{x}−\mathrm{6}} =\mathrm{5}^{−\frac{{x}^{\mathrm{3}} −\mathrm{10}{x}+\mathrm{3}}{{x}}} \\ $$$$\bullet\:\:{x}^{\mathrm{2}} −{x}−\mathrm{6}=\mathrm{0}\:\wedge\:−\frac{{x}^{\mathrm{3}} −\mathrm{10}{x}+\mathrm{3}}{{x}}=\mathrm{0} \\ $$$$\:\:\:\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)=\mathrm{0}\:\wedge\:{x}^{\mathrm{3}} −\mathrm{10}{x}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:{x}=\mathrm{3},−\mathrm{2}\:\wedge\:\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:{x}=\mathrm{3},−\mathrm{2}\:\wedge\:{x}=\mathrm{3},\frac{−\mathrm{3}\pm\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{3} \\ $$$$\bullet\left({x}^{\mathrm{2}} −{x}−\mathrm{6}\right)\mathrm{log3}=\left(−\frac{{x}^{\mathrm{3}} −\mathrm{10}{x}+\mathrm{3}}{{x}}\right)\mathrm{log5} \\ $$$$\:\:\:\:−\frac{{x}^{\mathrm{3}} −\mathrm{10}{x}+\mathrm{3}}{{x}\left({x}^{\mathrm{2}} −{x}−\mathrm{6}\right)}=\frac{\mathrm{log3}\:}{\mathrm{log5}}=\mathrm{log}_{\mathrm{5}} \mathrm{3} \\ $$$$\left(\mathrm{log}_{\mathrm{5}} \mathrm{3}+\mathrm{1}\right){x}^{\mathrm{3}} −\left(\mathrm{log}_{\mathrm{5}} \mathrm{3}\right){x}^{\mathrm{2}} −\mathrm{3}\left(\mathrm{2log}_{\mathrm{5}} \mathrm{3}+\mathrm{5}\right){x}+\mathrm{3}=\mathrm{0} \\ $$$$\: \\ $$$$… \\ $$

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