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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) )

2logax+1logaax+3logaa2x=02logax+1logaax+1logaa2x+2logaa2x=02(loga2x+logxlogx.loga2x)=−(loga2x+logaxlogax.loga2x)2loga2x2.logax=−logx.loga3x22(loga2+logx2).(loga+logx)=−logx(loga3+logx2)2(2+2logx)(1+logx)=−logx(3+2logx)4+8logx+4(logx)2=−3logx−2(logx)26(logx)2+11logx+4=0logx=−11±121−9612logx=−11±512=−1612,−612=−43,−12⇒x=a−43,a−12

Commented by NECx last updated on 30/Aug/17

thanks boss

thanksboss

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