maujanja au hujaelewa? Nway, ngoja niende internet cafe nikawaandikie njia vizuri.
Mmh! Mkuu hujafika tu huko internet cafe hadi leo utumwagie hako ka kipengele?
2^x =4x......log
log2 ^x=log4x.
Xlog2=log(2^2)X.
Xlog2=log(2^2)+logX.
Xlog2=2log2+logX
logX=Xlog2-2log2
logX=(X-2)Log2
X=(X-2)2
X=2X-4
4=2X-X
X=4.
Kumbe una ubavu katika mitaa hii...Hongera zako sana
2^x =4x......log
log2 ^x=log4x.
Xlog2=log(2^2)X.
Xlog2=log(2^2)+logX.
Xlog2=2log2+logX
logX=Xlog2-2log2
logX=(X-2)Log2
X=(X-2)2
X=2X-4
4=2X-X
X=4.
Acha kamba wewe! Umeondoa vipi logarithms hizo?2logX=(X-2)Log2
X=(X-2)2
2^x =4x......log
log2 ^x=log4x.
Xlog2=log(2^2)X.
Xlog2=log(2^2)+logX.
Xlog2=2log2+logX
logX=Xlog2-2log2
logX=(X-2)Log2
X=(X-2)2
X=2X-4
4=2X-X
X=4.
Haiwezi kuwa 0 baba
2[SUP]x[/SUP]=4X , 2[SUP]0[/SUP]=4(0), 2[SUP]0[/SUP]=1 and 4(0)=0 wapi na wapi.
naona nyota nyota tu hata sielewi.by they way taaluma yangu ni sheria.lol2^x =4x......log
log2 ^x=log4x.
Xlog2=log(2^2)X.
Xlog2=log(2^2)+logX.
Xlog2=2log2+logX
logX=Xlog2-2log2
logX=(X-2)Log2
X=(X-2)2
X=2X-4
4=2X-X
X=4.
Yupo sahihi jamani. Hebu rudieni post zake vizuri.