Find the value of x

[Mutta]

2^x = 4x( 2 power x is equal to 4x).Show your working clearly.Any one who can solve it help me.

 

[Preta]

wapi Kongosho.......mi mgeni sana hapa.....

 

[Kang]

 

[Kituko]

2^x = 4x( 2 power x is equal to 4x).Show your working clearly.Any one who can solve it help me.

1) Apply log ya base 10:
log2^x=log4x
xlog2=log4+logX
xlog2=log(4+x)
x=log(4+x)/log2
x=(4+x)/2
2x=4+x
x=4

 

[by default]

1) Apply log ya base 10:
log2^x=log4x
xlog2=log4+logX
xlog2=log(4+x)
x=log(4+x)/log2
x=(4+x)/2
2x=4+x
x=4

big up mkuu

 

[Chakunyuma]

1) Apply log ya base 10:
log2^x=log4x
xlog2=log4+logX
xlog2=log(4+x)
x=log(4+x)/log2
x=(4+x)/2
2x=4+x
x=4

Hapana mkuu cheki vizuri kwenye blue!Ipo hivi
2[SUP]x[/SUP]=4x
Apply log base ten
log(2[SUP]x[/SUP])=log(4x)
xlog2=log4 +logx
xlog2=log(2[SUP]2[/SUP]) +logx
xlog2=2log2 +logx
xlog2-2log2 =logx
(x-2)log2 =logx
x-2=logx/log2
x-2= x/2
2x-4 =x
2x-x =4
x=4
Umegundua kosa la kikanuni hapo?

 

[NingaR]

Hapana mkuu cheki vizuri kwenye blue!Ipo hivi
2[SUP]x[/SUP]=4x
Apply log base ten
log(2[SUP]x[/SUP])=log(4x)
xlog2=log4 +logx
xlog2=log(2[SUP]2[/SUP]) +logx
xlog2=2log2 +logx
xlog2-2log2 =logx
(x-2)log2 =logx
x-2=logx/log2
x-2= x/2
2x-4 =x
2x-x =4
x=4
Umegundua kosa la kikanuni hapo?

OYA logx/log2 ni logx base2 sio logx/2

 

[DASA]

Gonga hapa: https://www.jamiiforums.com/jukwaa-la-elimu-education-forum/178996-msaada-wa-haraka-hii-hesabu.html

 

[simplemind]

Problem has no exact solution. Can be solved graphically or by numerical methods.


Graphic solution:

Draw curve y=2^x and the line y= -4x . Point of intersection of curve and line is the required solution. (lies between x=-1 and x=0)

 

[Kang]

1) Apply log ya base 10:
log2^x=log4x
xlog2=log4+logX
xlog2=log(4+x)
x=log(4+x)/log2
x=(4+x)/2
2x=4+x
x=4

Hapana mkuu cheki vizuri kwenye blue!Ipo hivi
2[SUP]x[/SUP]=4x
Apply log base ten
log(2[SUP]x[/SUP])=log(4x)
xlog2=log4 +logx
xlog2=log(2[SUP]2[/SUP]) +logx
xlog2=2log2 +logx
xlog2-2log2 =logx
(x-2)log2 =logx
x-2=logx/log2
x-2= x/2
2x-4 =x
2x-x =4
x=4
Umegundua kosa la kikanuni hapo?

Hauwezi ku cancel logarithms kihivyo.

 

[DASA]

Ongezea na hii 3[SUP]X[/SUP]=9X

 

[King Kong III]

Haa mnanikumbusha mabanda yangu ya number.

 

[Husninyo]

Gonga hapa: https://www.jamiiforums.com/jukwaa-la-elimu-education-forum/178996-msaada-wa-haraka-hii-hesabu.html

DASA naona hesabu yako imejirudia.

 

[Mbimbinho]

Gonga hapa: https://www.jamiiforums.com/jukwaa-la-elimu-education-forum/178996-msaada-wa-haraka-hii-hesabu.html

DASA kutumia desa haimo mkuu..:nono:

 

[deogan]

Hapa naona tu kanuni mpya za log!

 

[Msafiri Kasian]

Dah! Jamani eeh! Mbona Kang ameshaelezea namna ya kutumia graph kusolve hyo hesabu? Log concept haipandi hapo jamani,mnaotumia log hapo ni Uongo unaofanana na Ukweli. Tumieni graph acheni uvivu watanzania,haya mabishano tunayoyaendeleza hapa,tulishabishana sana tukiwa shule,sasa tuuelewe ukweli tufanye kazi ya kuboresha maisha yetu.

 

[Msafiri Kasian]

Hauwezi ku cancel logarithms kihivyo.

Bg up kaka! Waambie hao,mi nilishaelewa tangu zamani,lakini humu watu hawataki kubadilika.

 

[Msafiri Kasian]

Hauwezi ku cancel logarithms kihivyo.

Bg up kaka Kang! Waambie hao,mi nilishaelewa tangu zamani,lakini humu watu hawataki kubadilika.

 

[DASA]

DASA naona hesabu yako imejirudia.

Niambie mtaalam, Ndio mana ikabidi niwape ule mtiririko ili wadadavue na kudadavulika wenyewe.

 

[Mutta]

1) Apply log ya base 10:
log2^x=log4x
xlog2=log4+logX
xlog2=log(4+x)
x=log(4+x)/log2
x=(4+x)/2
2x=4+x
x=4

hapo kwenye
x = l(4+x)/log2
x=4+x/2 sio kweli.there no such a law in logarithms