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(11628) (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.
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