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log2-X-log-2-2X-log-2-4X-X-




Question Number 130168 by Adel last updated on 23/Jan/21
log2_X ∙log 2_(2X) =log 2_(4X)             X=?
log2Xlog22X=log24XX=?
Commented by liberty last updated on 23/Jan/21
do you meant log _x (2).log _(2x) (2) = log _(4x) (2)?
doyoumeantlogx(2).log2x(2)=log4x(2)?
Answered by liberty last updated on 23/Jan/21
 (1/(log _2 (x))).(1/(log _2 (2x))) = (1/(log _2 (4x)))   2+log _2 (x)=log _2 (x)[ 1+log _2 (x) ]  let log _2 (x)=p  ⇒2+p = p+p^2  ; p^2 =2 → { ((p=(√2))),((p=−(√2))) :}   ⇒log _2 (x)=± (√2)  ⇒x = 2^(± (√2))  ; x =  { (2^(√2) ),((1/2^(√2) )) :}
1log2(x).1log2(2x)=1log2(4x)2+log2(x)=log2(x)[1+log2(x)]letlog2(x)=p2+p=p+p2;p2=2{p=2p=2log2(x)=±2x=2±2;x={22122
Commented by Adel last updated on 23/Jan/21
thanks sir
thankssir
Answered by MJS_new last updated on 23/Jan/21
((ln 2)/(ln x))×((ln 2)/(ln 2x))=((ln 2)/(ln 4x))  t=ln x  ((ln 2)/t)×((ln 2)/(t+ln 2))=((ln 2)/(t+2ln 2))  t(t+ln 2)=ln 2 (t+2ln 2)  t^2 =2(ln 2)^2   t=±(√2)ln 2  x=2^(−(√2)) ∨x=2^(√2)
ln2lnx×ln2ln2x=ln2ln4xt=lnxln2t×ln2t+ln2=ln2t+2ln2t(t+ln2)=ln2(t+2ln2)t2=2(ln2)2t=±2ln2x=22x=22
Commented by liberty last updated on 23/Jan/21
yes...my answer correct..
yesmyanswercorrect..
Commented by Adel last updated on 23/Jan/21
tanks sir
tankssir

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