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Question-18123

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Question Number 18123 by mondodotto@gmail.com last updated on 15/Jul/17
Commented by prakash jain last updated on 15/Jul/17
log (x−3)^(1/3) +log 5−log (x−2)^2 = log 5=log (x−2)^2 −log (x−3)^(1/3) 5=(((x−2)^2 )/((x−3)^(1/3) )) 5(x−3)^(1/3) =(x−2)^2 125(x−3)=(x−2)^6 x=2+u u^6 =125u−125 u^6 −125u+125=0 one root will near 1 (1+v)^6 −125(1+v)+125=0 v=(1/(119))⇒v=.008⇒u=1.0084 x=3.0084 −−−−−−−−−− 2^6 −125×2+125<0 3^6 −125×3+125>0 root between 2 and 3 2.5 (5^6 /2^6 )−((125×5)/2)+125=125−160+64=29 root is below 2.5 but a large difference ((5/2)−v)^6 −((125×5)/2)+125v+125=0 (5^6 /2^6 )(1−((2v)/5))^6 −((125×5)/2)+125v+125=0 5^3 (1−((12v)/5))−160+64v+64=0 125−300v−160+64v+64=0 −236v+29=0⇒v=((29)/(236))=.12 v=.12 is large difference so we we can compute more accurate value of v. 5^3 (1−((12v)/5)+((6×5)/2)×((2^2 v^2 )/5^2 ))−160+64v+64=0 300v^2 −236v+29=0 v=.15 u=2.35 x=4.35
$$\mathrm{log}\:\left({x}−\mathrm{3}\right)^{\mathrm{1}/\mathrm{3}} +\mathrm{log}\:\mathrm{5}−\mathrm{log}\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} = \\ $$$$\mathrm{log}\:\mathrm{5}=\mathrm{log}\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{log}\:\left({x}−\mathrm{3}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{5}=\frac{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\left({x}−\mathrm{3}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$$\mathrm{5}\left({x}−\mathrm{3}\right)^{\mathrm{1}/\mathrm{3}} =\left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{125}\left({x}−\mathrm{3}\right)=\left({x}−\mathrm{2}\right)^{\mathrm{6}} \\ $$$${x}=\mathrm{2}+{u} \\ $$$${u}^{\mathrm{6}} =\mathrm{125}{u}−\mathrm{125} \\ $$$${u}^{\mathrm{6}} −\mathrm{125}{u}+\mathrm{125}=\mathrm{0} \\ $$$$\mathrm{one}\:\mathrm{root}\:\mathrm{will}\:\mathrm{near}\:\mathrm{1} \\ $$$$\left(\mathrm{1}+{v}\right)^{\mathrm{6}} −\mathrm{125}\left(\mathrm{1}+{v}\right)+\mathrm{125}=\mathrm{0} \\ $$$${v}=\frac{\mathrm{1}}{\mathrm{119}}\Rightarrow{v}=.\mathrm{008}\Rightarrow{u}=\mathrm{1}.\mathrm{0084} \\ $$$${x}=\mathrm{3}.\mathrm{0084} \\ $$$$−−−−−−−−−− \\ $$$$\mathrm{2}^{\mathrm{6}} −\mathrm{125}×\mathrm{2}+\mathrm{125}<\mathrm{0} \\ $$$$\mathrm{3}^{\mathrm{6}} −\mathrm{125}×\mathrm{3}+\mathrm{125}>\mathrm{0} \\ $$$$\mathrm{root}\:\mathrm{between}\:\mathrm{2}\:\mathrm{and}\:\mathrm{3} \\ $$$$\mathrm{2}.\mathrm{5} \\ $$$$\frac{\mathrm{5}^{\mathrm{6}} }{\mathrm{2}^{\mathrm{6}} }−\frac{\mathrm{125}×\mathrm{5}}{\mathrm{2}}+\mathrm{125}=\mathrm{125}−\mathrm{160}+\mathrm{64}=\mathrm{29} \\ $$$$\mathrm{root}\:\mathrm{is}\:\mathrm{below}\:\mathrm{2}.\mathrm{5}\:\mathrm{but}\:\mathrm{a}\:\mathrm{large}\:\mathrm{difference} \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{2}}−{v}\right)^{\mathrm{6}} −\frac{\mathrm{125}×\mathrm{5}}{\mathrm{2}}+\mathrm{125}{v}+\mathrm{125}=\mathrm{0} \\ $$$$\frac{\mathrm{5}^{\mathrm{6}} }{\mathrm{2}^{\mathrm{6}} }\left(\mathrm{1}−\frac{\mathrm{2}{v}}{\mathrm{5}}\right)^{\mathrm{6}} −\frac{\mathrm{125}×\mathrm{5}}{\mathrm{2}}+\mathrm{125}{v}+\mathrm{125}=\mathrm{0} \\ $$$$\mathrm{5}^{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{12}{v}}{\mathrm{5}}\right)−\mathrm{160}+\mathrm{64}{v}+\mathrm{64}=\mathrm{0} \\ $$$$\mathrm{125}−\mathrm{300}{v}−\mathrm{160}+\mathrm{64}{v}+\mathrm{64}=\mathrm{0} \\ $$$$−\mathrm{236}{v}+\mathrm{29}=\mathrm{0}\Rightarrow{v}=\frac{\mathrm{29}}{\mathrm{236}}=.\mathrm{12} \\ $$$${v}=.\mathrm{12}\:\mathrm{is}\:\mathrm{large}\:\mathrm{difference}\:\mathrm{so}\:\mathrm{we} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{compute}\:\mathrm{more}\:\mathrm{accurate} \\ $$$$\mathrm{value}\:\mathrm{of}\:{v}. \\ $$$$\mathrm{5}^{\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{12}{v}}{\mathrm{5}}+\frac{\mathrm{6}×\mathrm{5}}{\mathrm{2}}×\frac{\mathrm{2}^{\mathrm{2}} {v}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} }\right)−\mathrm{160}+\mathrm{64}{v}+\mathrm{64}=\mathrm{0} \\ $$$$\mathrm{300}{v}^{\mathrm{2}} −\mathrm{236}{v}+\mathrm{29}=\mathrm{0} \\ $$$${v}=.\mathrm{15} \\ $$$${u}=\mathrm{2}.\mathrm{35} \\ $$$${x}=\mathrm{4}.\mathrm{35} \\ $$
Commented by mondodotto@gmail.com last updated on 15/Jul/17
thanx a lot prakash
$$\mathrm{thanx}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{prakash} \\ $$

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