Question-20626

Question Number 20626 by NECx last updated on 29/Aug/17

Answered by $@ty@m last updated on 30/Aug/17

(2/(log_a x))+(1/(log_a ax))+(3/(log_a a^2 x))=0 (2/(log_a x))+(1/(log_a ax))+(1/(log_a a^2 x))+(2/(log_a a^2 x))=0 2(((loga^2 x+logx)/(logx.loga^2 x)))=−(((loga^2 x+logax)/(logax.loga^2 x))) 2loga^2 x^2 .logax=−logx.loga^3 x^2 2(loga^2 +logx^2 ).(loga+logx)=−logx(loga^3 +logx^2 ) 2(2+2logx)(1+logx)=−logx(3+2logx) 4+8logx+4(logx)^2 =−3logx−2(logx)^2 6(logx)^2 +11logx+4=0 logx=((−11±(√(121−96)))/(12)) logx=((−11±5)/(12))=−((16)/(12)), −(6/(12))=−(4/3), −(1/2) ⇒x=a^((−4)/3) , a^(((−1)/2) )

$$\frac{\mathrm{2}}{{log}_{{a}} {x}}+\frac{\mathrm{1}}{{log}_{{a}} {ax}}+\frac{\mathrm{3}}{{log}_{{a}} {a}^{\mathrm{2}} {x}}=\mathrm{0} \\ $$$$\frac{\mathrm{2}}{{log}_{{a}} {x}}+\frac{\mathrm{1}}{{log}_{{a}} {ax}}+\frac{\mathrm{1}}{{log}_{{a}} {a}^{\mathrm{2}} {x}}+\frac{\mathrm{2}}{{log}_{{a}} {a}^{\mathrm{2}} {x}}=\mathrm{0} \\ $$$$\mathrm{2}\left(\frac{{loga}^{\mathrm{2}} {x}+{logx}}{{logx}.{loga}^{\mathrm{2}} {x}}\right)=−\left(\frac{{loga}^{\mathrm{2}} {x}+{logax}}{{logax}.{loga}^{\mathrm{2}} {x}}\right) \\ $$$$\mathrm{2}{loga}^{\mathrm{2}} {x}^{\mathrm{2}} .{logax}=−{logx}.{loga}^{\mathrm{3}} {x}^{\mathrm{2}} \\ $$$$\mathrm{2}\left({loga}^{\mathrm{2}} +{logx}^{\mathrm{2}} \right).\left({loga}+{logx}\right)=−{logx}\left({loga}^{\mathrm{3}} +{logx}^{\mathrm{2}} \right) \\ $$$$\mathrm{2}\left(\mathrm{2}+\mathrm{2}{logx}\right)\left(\mathrm{1}+{logx}\right)=−{logx}\left(\mathrm{3}+\mathrm{2}{logx}\right) \\ $$$$\mathrm{4}+\mathrm{8}{logx}+\mathrm{4}\left({logx}\right)^{\mathrm{2}} =−\mathrm{3}{logx}−\mathrm{2}\left({logx}\right)^{\mathrm{2}} \\ $$$$\mathrm{6}\left({logx}\right)^{\mathrm{2}} +\mathrm{11}{logx}+\mathrm{4}=\mathrm{0} \\ $$$${logx}=\frac{−\mathrm{11}\pm\sqrt{\mathrm{121}−\mathrm{96}}}{\mathrm{12}} \\ $$$${logx}=\frac{−\mathrm{11}\pm\mathrm{5}}{\mathrm{12}}=−\frac{\mathrm{16}}{\mathrm{12}},\:−\frac{\mathrm{6}}{\mathrm{12}}=−\frac{\mathrm{4}}{\mathrm{3}},\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}={a}^{\frac{−\mathrm{4}}{\mathrm{3}}} ,\:{a}^{\frac{−\mathrm{1}}{\mathrm{2}}\:} \\ $$$$ \\ $$

Commented by NECx last updated on 30/Aug/17

$${thanks}\:{boss} \\ $$

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