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Question Number 198809 by dimentri last updated on 24/Oct/23
sum of roots log _3 x + log _3 (2,5) + log _x 9 = 3+ log _x 5
$$\:\:\:\:\:{sum}\:{of}\:{roots}\: \\ $$$$\:\mathrm{log}\:_{\mathrm{3}} {x}\:+\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2},\mathrm{5}\right)\:+\:\mathrm{log}\:_{{x}} \mathrm{9}\:=\:\mathrm{3}+\:\mathrm{log}\:_{{x}} \mathrm{5}\: \\ $$
Answered by a.lgnaoui last updated on 24/Oct/23
((logx)/(log3))+((log(2,5))/(log3))+((log9)/(logx))=3+((log5)/(logx)) (logx)^2 +logx.log(2,5)+2(log3)^2 =3log3logx+log3.log5 posons logx=z z^2 +z[log(2,5)−3log3]+log3(2log3−log5)=0 △=[log(2,5)−3log3]^2 −4log3(2log3−log5) log(2,5)=log(((25)/(10)))=log25−1=2log5−1 z^2 +z(2log5−3log3−1)+log3(2log3−log5)=0 △=(2log5−3log3−1)^2 −8(log3)^2 + 4log3.log5 =1+(log3)^2 −16log3log5+6log3 +4(log5)^2 −4log5 =(2log5−1)^2 +(log3−8log5)^2 =+6log3−64(log5)^2 =−6log5−1)(10log5−1)+ △=6log3−(10log5−1)(6log5+1)+ (8log5−log3)^2 z=(((3log3+1−2log5)±(√(6log3−(6log5+1)(10log5−1)+(8log5−log3)^2 )))/2) x=e^((3log3−2log5+1)/2) ×e^(±(√(6log3+(8log5−3)^2 −6(log5−1)(6kog5+1))) /2) ⇒Sum of roots is r=e^((3/2)log3−log5+(1/2))
$$\frac{\mathrm{logx}}{\mathrm{log3}}+\frac{\mathrm{log}\left(\mathrm{2},\mathrm{5}\right)}{\mathrm{log3}}+\frac{\mathrm{log9}}{\mathrm{logx}}=\mathrm{3}+\frac{\mathrm{log5}}{\mathrm{logx}} \\ $$$$\left(\mathrm{logx}\right)^{\mathrm{2}} +\mathrm{logx}.\mathrm{log}\left(\mathrm{2},\mathrm{5}\right)+\mathrm{2}\left(\mathrm{log3}\right)^{\mathrm{2}} \\ $$$$=\mathrm{3log3logx}+\mathrm{log3}.\mathrm{log5} \\ $$$$\mathrm{posons}\:\:\mathrm{logx}=\boldsymbol{\mathrm{z}} \\ $$$$\boldsymbol{\mathrm{z}}^{\mathrm{2}} +\boldsymbol{\mathrm{z}}\left[\boldsymbol{\mathrm{log}}\left(\mathrm{2},\mathrm{5}\right)−\mathrm{3}\boldsymbol{\mathrm{log}}\mathrm{3}\right]+\boldsymbol{\mathrm{log}}\mathrm{3}\left(\mathrm{2}\boldsymbol{\mathrm{log}}\mathrm{3}−\boldsymbol{\mathrm{log}}\mathrm{5}\right)=\mathrm{0} \\ $$$$ \\ $$$$\bigtriangleup=\left[\boldsymbol{\mathrm{log}}\left(\mathrm{2},\mathrm{5}\right)−\mathrm{3}\boldsymbol{\mathrm{log}}\mathrm{3}\right]^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{log}}\mathrm{3}\left(\mathrm{2}\boldsymbol{\mathrm{log}}\mathrm{3}−\boldsymbol{\mathrm{log}}\mathrm{5}\right) \\ $$$$\:\:\boldsymbol{\mathrm{log}}\left(\mathrm{2},\mathrm{5}\right)=\mathrm{log}\left(\frac{\mathrm{25}}{\mathrm{10}}\right)=\mathrm{log25}−\mathrm{1}=\mathrm{2log5}−\mathrm{1} \\ $$$$\boldsymbol{\mathrm{z}}^{\mathrm{2}} +\boldsymbol{\mathrm{z}}\left(\mathrm{2}\boldsymbol{\mathrm{log}}\mathrm{5}−\mathrm{3}\boldsymbol{\mathrm{log}}\mathrm{3}−\mathrm{1}\right)+\boldsymbol{\mathrm{log}}\mathrm{3}\left(\mathrm{2}\boldsymbol{\mathrm{log}}\mathrm{3}−\boldsymbol{\mathrm{log}}\mathrm{5}\right)=\mathrm{0} \\ $$$$ \\ $$$$\bigtriangleup=\left(\mathrm{2log5}−\mathrm{3log3}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{8}\left(\mathrm{log3}\right)^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4log3}.\mathrm{log5} \\ $$$$=\mathrm{1}+\left(\mathrm{log3}\right)^{\mathrm{2}} −\mathrm{16}\boldsymbol{\mathrm{log}}\mathrm{3}\boldsymbol{\mathrm{log}}\mathrm{5}+\mathrm{6}\boldsymbol{\mathrm{log}}\mathrm{3} \\ $$$$\:\:\:+\mathrm{4}\left(\boldsymbol{\mathrm{log}}\mathrm{5}\right)^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{log}}\mathrm{5} \\ $$$$=\left(\mathrm{2}\boldsymbol{\mathrm{log}}\mathrm{5}−\mathrm{1}\right)^{\mathrm{2}} +\left(\boldsymbol{\mathrm{log}}\mathrm{3}−\mathrm{8}\boldsymbol{\mathrm{log}}\mathrm{5}\right)^{\mathrm{2}} \\ $$$$=+\mathrm{6log3}−\mathrm{64}\left(\mathrm{log5}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left.=−\mathrm{6}\boldsymbol{\mathrm{log}}\mathrm{5}−\mathrm{1}\right)\left(\mathrm{10}\boldsymbol{\mathrm{log}}\mathrm{5}−\mathrm{1}\right)+ \\ $$$$ \\ $$$$\:\:\bigtriangleup=\mathrm{6}\boldsymbol{\mathrm{log}}\mathrm{3}−\left(\mathrm{10}\boldsymbol{\mathrm{log}}\mathrm{5}−\mathrm{1}\right)\left(\mathrm{6}\boldsymbol{\mathrm{log}}\mathrm{5}+\mathrm{1}\right)+ \\ $$$$\:\:\:\:\:\:\left(\mathrm{8}\boldsymbol{\mathrm{log}}\mathrm{5}−\boldsymbol{\mathrm{log}}\mathrm{3}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{z}}=\frac{\left(\mathrm{3}\boldsymbol{\mathrm{log}}\mathrm{3}+\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{log}}\mathrm{5}\right)\pm\sqrt{\mathrm{6}\boldsymbol{\mathrm{log}}\mathrm{3}−\left(\mathrm{6}\boldsymbol{\mathrm{log}}\mathrm{5}+\mathrm{1}\right)\left(\mathrm{10}\boldsymbol{\mathrm{log}}\mathrm{5}−\mathrm{1}\right)+\left(\mathrm{8}\boldsymbol{\mathrm{log}}\mathrm{5}−\boldsymbol{\mathrm{log}}\mathrm{3}\right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$ \\ $$$$\:\: \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{e}}^{\left(\mathrm{3}\boldsymbol{\mathrm{log}}\mathrm{3}−\mathrm{2}\boldsymbol{\mathrm{log}}\mathrm{5}+\mathrm{1}\right)/\mathrm{2}} ×\mathrm{e}^{\pm\sqrt{\mathrm{6log3}+\left(\mathrm{8log5}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{6}\left(\mathrm{log5}−\mathrm{1}\right)\left(\mathrm{6kog5}+\mathrm{1}\right)}\:/\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\boldsymbol{\mathrm{Sum}}\:\:\:\mathrm{of}\:\boldsymbol{\mathrm{roots}}\:\boldsymbol{\mathrm{is}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{r}}=\boldsymbol{\mathrm{e}}^{\frac{\mathrm{3}}{\mathrm{2}}\boldsymbol{\mathrm{log}}\mathrm{3}−\boldsymbol{\mathrm{log}}\mathrm{5}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$
Commented by mr W last updated on 24/Oct/23
wrong! x=e^(a±b) ⇒Σx=e^a (e^b +e^(−b) )≠e^a
$${wrong}! \\ $$$${x}={e}^{{a}\pm{b}} \\ $$$$\Rightarrow\Sigma{x}={e}^{{a}} \left({e}^{{b}} +{e}^{−{b}} \right)\neq{e}^{{a}} \\ $$
Commented by a.lgnaoui last updated on 24/Oct/23
yes thanks i use logaritme base 10 i think that my answer wil be exact if the we change log by ln in the given equation.(c′ est un piege!)
$$\mathrm{yes}\:\mathrm{thanks} \\ $$$$\mathrm{i}\:\mathrm{use}\:\mathrm{logaritme}\:\mathrm{base}\:\mathrm{10}\: \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{that}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{wil}\:\mathrm{be}\:\mathrm{exact} \\ $$$$\:\mathrm{if}\:\mathrm{the}\:\mathrm{we}\:\mathrm{change}\:\mathrm{log}\:\mathrm{by}\:\:\mathrm{ln}\:\mathrm{in}\:\mathrm{the} \\ $$$$\:\mathrm{given}\:\mathrm{equation}.\left(\mathrm{c}'\:\mathrm{est}\:\mathrm{un}\:\mathrm{piege}!\right) \\ $$$$ \\ $$
Commented by mr W last updated on 24/Oct/23
oui!
$${oui}! \\ $$
Answered by mr W last updated on 24/Oct/23
log _3 x + log _3 (2,5)−log_3 3^3 = log _x 5−log_x 9 log_3 ((2.5x)/(27))=log_x (5/9) ((ln x+ln (((2.5)/(27))))/(log 3))=((ln (5/9))/(ln x)) (ln x)(ln x+ln ((2.5)/(27)))=(ln 3)(ln (5/9)) let t=ln x t^2 +(ln ((2.5)/(27)))t−(ln 3)(ln (5/9))=0 t_(1,2) =(1/2)[−ln ((2.5)/(27))±(√((ln ((2.5)/(27)))^2 +4(ln 3)(ln (5/9))))] t_1 +t_2 =−ln ((2.5)/(27))=ln ((54)/5) x_(1,2) =e^t_(1,2) sum of roots: x_1 +x_2 =e^t_1 +e^t_2 =.... product of roots: x_1 x_2 =e^t_1 e^t_2 =e^(t_1 +t_2 ) =e^(ln ((54)/5)) =((54)/5)
$$\:\mathrm{log}\:_{\mathrm{3}} {x}\:+\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2},\mathrm{5}\right)−\mathrm{log}_{\mathrm{3}} \:\mathrm{3}^{\mathrm{3}} =\:\mathrm{log}\:_{{x}} \mathrm{5}−\mathrm{log}_{{x}} \:\mathrm{9} \\ $$$$\mathrm{log}_{\mathrm{3}} \:\frac{\mathrm{2}.\mathrm{5}{x}}{\mathrm{27}}=\mathrm{log}_{{x}} \:\frac{\mathrm{5}}{\mathrm{9}} \\ $$$$\frac{\mathrm{ln}\:{x}+\mathrm{ln}\:\left(\frac{\mathrm{2}.\mathrm{5}}{\mathrm{27}}\right)}{\mathrm{log}\:\mathrm{3}}=\frac{\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{9}}}{\mathrm{ln}\:{x}} \\ $$$$\left(\mathrm{ln}\:{x}\right)\left(\mathrm{ln}\:{x}+\mathrm{ln}\:\frac{\mathrm{2}.\mathrm{5}}{\mathrm{27}}\right)=\left(\mathrm{ln}\:\mathrm{3}\right)\left(\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{9}}\right) \\ $$$${let}\:{t}=\mathrm{ln}\:{x} \\ $$$${t}^{\mathrm{2}} +\left(\mathrm{ln}\:\frac{\mathrm{2}.\mathrm{5}}{\mathrm{27}}\right){t}−\left(\mathrm{ln}\:\mathrm{3}\right)\left(\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{9}}\right)=\mathrm{0} \\ $$$${t}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left[−\mathrm{ln}\:\frac{\mathrm{2}.\mathrm{5}}{\mathrm{27}}\pm\sqrt{\left(\mathrm{ln}\:\frac{\mathrm{2}.\mathrm{5}}{\mathrm{27}}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{ln}\:\mathrm{3}\right)\left(\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{9}}\right)}\right] \\ $$$${t}_{\mathrm{1}} +{t}_{\mathrm{2}} =−\mathrm{ln}\:\frac{\mathrm{2}.\mathrm{5}}{\mathrm{27}}=\mathrm{ln}\:\frac{\mathrm{54}}{\mathrm{5}} \\ $$$${x}_{\mathrm{1},\mathrm{2}} ={e}^{{t}_{\mathrm{1},\mathrm{2}} } \\ $$$$\underline{{sum}\:{of}\:{roots}:} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} ={e}^{{t}_{\mathrm{1}} } +{e}^{{t}_{\mathrm{2}} } =…. \\ $$$$ \\ $$$$\underline{{product}\:{of}\:{roots}:} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} ={e}^{{t}_{\mathrm{1}} } {e}^{{t}_{\mathrm{2}} } ={e}^{{t}_{\mathrm{1}} +{t}_{\mathrm{2}} } ={e}^{\mathrm{ln}\:\frac{\mathrm{54}}{\mathrm{5}}} =\frac{\mathrm{54}}{\mathrm{5}} \\ $$

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