# SOLUTION: The question is: log(base x)2 + log(base2)x= -2 What I did: log(base x)2 + log(base2)x= -2 log(base x)2+ 1/log(base x)2= -2 (log (base x)2)^2 + 1= -2log(base x)2 Take log(b

Algebra ->  Algebra  -> Exponential-and-logarithmic-functions -> SOLUTION: The question is: log(base x)2 + log(base2)x= -2 What I did: log(base x)2 + log(base2)x= -2 log(base x)2+ 1/log(base x)2= -2 (log (base x)2)^2 + 1= -2log(base x)2 Take log(b      Log On

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 Click here to see ALL problems on Exponential-and-logarithmic-functions Question 581749: The question is: log(base x)2 + log(base2)x= -2 What I did: log(base x)2 + log(base2)x= -2 log(base x)2+ 1/log(base x)2= -2 (log (base x)2)^2 + 1= -2log(base x)2 Take log(base x)2 as a: a^2 + 1= -2a a^2 + 2a + 1= 0 (a+1)(a+1)=0 a+1=0 a=-1 log(base x)2= -1 x^-1= 2 x= 1/2 I tried doing this, but I'm not sure if I'm right =) Thanks in advance!!Answer by lwsshak3(6469)   (Show Source): You can put this solution on YOUR website!log(base x)2 + log(base2)x= -2 convert to base 10 log2/logx+logx/log2=-2 LCD:log2logx log2^2+logx^2=-2log2logx logx^2+2log2logx+log2^2=0 let u=logx u^2=logx^2 Using quadratic formula: a=1, b=2log2, c=log2^2 u={-2log2±√[4log2^2-4*1*log2^2]}/2*1 u=[-2log2±√(4log2^2-4log2^2)]/2 u=[-2log2±√(0)]/2 u=-2log2/2 u=-log2=logx log(1/2)=log2^-1=-1*log2=-log2 x=1/2 I got the same answer so we may be doing it right. I had some difficulty following your computations, but your method is probably ok.