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Question Number 42627 by mondodotto@gmail.com last updated on 29/Aug/18

solve for x  (1/3)log(x−3)+log5−log(x−2)^2 =0

$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{\mathrm{log}}\left(\boldsymbol{{x}}−\mathrm{3}\right)+\boldsymbol{\mathrm{log}}\mathrm{5}−\boldsymbol{\mathrm{log}}\left(\boldsymbol{{x}}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$

Answered by $@ty@m last updated on 30/Aug/18

log (x−3)+3log 5−3log (x−2)^2 =log 1  ((125(x−3))/((x−2)^6 ))=1  (x−2)^6 =125x−375  x^6 −6x^5 .2+15x^4 .2^2 −20x^3 .2^3                +15x^2 .2^4 −6x.2^5 +2^6 =125x−375  x^6 −12x^5 +60x^4 −160x^3 +240x^2           −317x+439=0  to be continue...

$$\mathrm{log}\:\left({x}−\mathrm{3}\right)+\mathrm{3log}\:\mathrm{5}−\mathrm{3log}\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{log}\:\mathrm{1} \\ $$$$\frac{\mathrm{125}\left({x}−\mathrm{3}\right)}{\left({x}−\mathrm{2}\right)^{\mathrm{6}} }=\mathrm{1} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{6}} =\mathrm{125}{x}−\mathrm{375} \\ $$$${x}^{\mathrm{6}} −\mathrm{6}{x}^{\mathrm{5}} .\mathrm{2}+\mathrm{15}{x}^{\mathrm{4}} .\mathrm{2}^{\mathrm{2}} −\mathrm{20}{x}^{\mathrm{3}} .\mathrm{2}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{15}{x}^{\mathrm{2}} .\mathrm{2}^{\mathrm{4}} −\mathrm{6}{x}.\mathrm{2}^{\mathrm{5}} +\mathrm{2}^{\mathrm{6}} =\mathrm{125}{x}−\mathrm{375} \\ $$$${x}^{\mathrm{6}} −\mathrm{12}{x}^{\mathrm{5}} +\mathrm{60}{x}^{\mathrm{4}} −\mathrm{160}{x}^{\mathrm{3}} +\mathrm{240}{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:−\mathrm{317}{x}+\mathrm{439}=\mathrm{0} \\ $$$${to}\:{be}\:{continue}... \\ $$

Answered by ajfour last updated on 30/Aug/18

 ((125(x−3))/((x−2)^6 )) = 1  125(x−3)−(x−2)^6  = 0  ⇒  x ≈ 3.008  ,   x ≈ 4.351 .

$$\:\frac{\mathrm{125}\left({x}−\mathrm{3}\right)}{\left({x}−\mathrm{2}\right)^{\mathrm{6}} }\:=\:\mathrm{1} \\ $$$$\mathrm{125}\left({x}−\mathrm{3}\right)−\left({x}−\mathrm{2}\right)^{\mathrm{6}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:{x}\:\approx\:\mathrm{3}.\mathrm{008}\:\:,\:\:\:{x}\:\approx\:\mathrm{4}.\mathrm{351}\:. \\ $$

Commented by mondodotto@gmail.com last updated on 30/Aug/18

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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