Msaada wa haraka hii hesabu!!!

[Kalambo Junior]

2^x=4x apply log each side log2^x=log4x xlog2=log4+logx xlog2=2log2+logx xlog2-2log2=logx log2(x-2)=logx dividing by log2(x-2) throughout the eqn we obtain 1=logx/log2(x-2) this can also be writen as logxbase2(x-2)=1, into exponential form it will be 2(x-2)^1=x 2(x-2)=x 2x-4=x x=4 ok!

yes! it's correct

 

[Husninyo]

wakuu wote mnaohoji log ilivyopotea nilishasawazisha kwenye post yangu iliyofatia labda ilikuwa ngumu kuiona. fata mtiririko huu ili tuelewane
2[SUP]x[/SUP] =4x..............................................apply log
log2[SUP]x[/SUP]=log4x.
Xlog2=log(2[SUP]2[/SUP])X.
Xlog2=log(2[SUP]2[/SUP])+logX.
Xlog2=2log2+logX
logX=Xlog2-2log2
logX=(X-2)Log2.........................gawanya kwa (x-2)log2both sides
logx =1
(x-2)log2
logx =1.............................................................andika hiyo katika mfumo wa log under base,
log2(x-2)

log [SUB]2(x-2)[/SUB]X=1..................................................................................ibadilishe katika exponential form
2(x-2)[SUP]1[/SUP]=x
2x-4=x
2x-x=4
X=4.

 

[Anheuser]

logx =1
log2(x-2)

log [SUB]2(x-2)[/SUB]X=1..........................

Hii nayo ni kamba!

Kwenye numerator ume introduce base ya 2(x-2), ukapata log [SUB]2(x-2)[/SUB]X , chini ukaipotezea denominator!

Denominator nzima ya [x-2]log2 imeenda wapi?

 

[Husninyo]

Hii nayo ni kamba!

Kwenye numerator ume introduce base ya 2(x-2), ukapata log [SUB]2(x-2)[/SUB]X , chini ukaipotezea denominator!

Denominator nzima ya [x-2]log2 imeenda wapi?

ama kweli walimu wanakazi nzito. unakubali kama a*b=b*a?

 

[JF2050]

kaka upo sahihi hapo. Aisee hesabu ukiitupa na yenyewe inakutupa.
Fanya hivi, ukifika kwenye;
logx=log2(x-2) gawanya kwa log2(x-2) kila upande.
Utapata logx/log2(x-2)=1
hiyo log iandike katika mfumo wa log ambapo base yake itakuwa ni 2(x-2) halafu ibadilishe katika exponential form. Utapata jibu hapo.

Congratulations Husninyo!

 

[Anheuser]

ama kweli walimu wanakazi nzito. unakubali kama a*b=b*a?

a*b=b*a,

yes, but

(x-2)log2 is not equal to log2(x-2), nakataa!

the correct commutative re-statement of function (x-2)log2 is (log2)x-2.

Hilo kosa la kwanza.

logx =1.............................................................andika hiyo katika mfumo wa log under base,
log2(x-2)

Blunder la pili, unaposema "andika hiyo katika mfumo wa log under base"

inatakiwa iwe hivi:

If log x = 1
(log2)( x-2)

Then

2[SUP](x-2)log2[/SUP] = X (note that the base is 2, ambayo wewe after all huku specify unachukua base gani).

Which simplifies to:

2[SUP](x-2)(1) [/SUP]= X (kwa sababu log[SUB] 2[/SUB] 2 = 1)

Thus:

2[SUP]x [/SUP] / 2[SUP]2 [/SUP]= X

2[SUP]x [/SUP]= 4X

ambapo ndipo tulipoanzia, unarudi pale pale! Kwa hiyo approach yako ni mbovu na ya kisanii!

Eti "walimu wana kazi nzito"! Wewe mwalimu mwongo huwezi kunifundisha mimi.

 

[DASA]

a*b=b*a,

yes, but

(x-2)log2 is not equal to log2(x-2), nakataa!

the correct commutative re-statement of function (x-2)log2 is (log2)x-2.

Hilo kosa la kwanza.



Blunder la pili, unaposema "andika hiyo katika mfumo wa log under base"

inatakiwa iwe hivi:

If log x = 1
(log2)( x-2)

Then

2[SUP](x-2)log2[/SUP] = X (note that the base is 2, ambayo wewe after all huku specify unachukua base gani).

Which simplifies to:

2[SUP](x-2)(1) [/SUP]= X (kwa sababu log[SUB] 2[/SUB] 2 = 1)

Thus:

2[SUP]x [/SUP] / 2[SUP]2 [/SUP]= X

2[SUP]x [/SUP]= 4X

ambapo ndipo tulipoanzia, unarudi pale pale! Kwa hiyo approach yako ni mbovu na ya kisanii!

Eti "walimu wana kazi nzito"! Wewe mwalimu mwongo huwezi kunifundisha mimi.

Mkuu we umeshindikana. Hebu kajikumbushekumbushe kidogo vitabu vya huko nyuma.

Hiyo case ya kwanza fanya hivi basi (x-2)(Log2)=(Log2)(x-2)

Hiyo case ya pili kumbuka Loga/logb=Log[SUB]b[/SUB]a

 

[Husninyo]

Mkuu we umeshindikana. Hebu kajikumbushekumbushe kidogo vitabu vya huko nyuma.

Hiyo case ya kwanza fanya hivi basi (x-2)(Log2)=(Log2)(x-2)

Hiyo case ya pili kumbuka Loga/logb=Log[SUB]b[/SUB]a

dasa thank you. jamaa utafikiri ameambiwa tupo kwenye mashindano hapa. bora umenisaida kumjibu maana nilikuwa nishachoka.

 

[DASA]

dasa thank you. jamaa utafikiri ameambiwa tupo kwenye mashindano hapa. bora umenisaida kumjibu maana nilikuwa nishachoka.

Pole sana dadangu. hapo nadhani atakubaliana sasa.

 

[Husninyo]

Pole sana dadangu. hapo nadhani atakubaliana sasa.

hahhahaha, watu wabishi. kuna mtu hadi alilamba kinyesi ili aamini kama kweli ni kinyesi.

 

[DASA]

hahhahaha, watu wabishi. kuna mtu hadi alilamba kinyesi ili aamini kama kweli ni kinyesi.

Hahahahahaha....., hiyo kali. Ngoja tuone kama watabisha tena.

 

[HAMY-D]

2^x=4x apply log
xlog2=log 4x
xlog 2=log 4+log x
xlog 2-2log 2=log x
log 2 (x-2)=log x cancel log throughout
2(x-2)=x
2x-4=x
x=4 ,
hili swali lipo kama 3^x= 9x , jibu ni x=3.

 

[Billie]

Nimepitia mabishano na solution zenu pamoja na thread ya huyu jamaa nikiwa na T.O wa advanced mathematics mwaka 2009 ametoka PUGU na kwa sasa yupo UDSm anasoma BCOM finance Tumecheka sana kutokana kuwa hili swali tumeshalifanyia kazi na huyu T.O ELINEEMA MOLA alipata 98% ya Pure math form six.
JIBU LA SWALI KUNA JAMAA MMOJA kapata ameandika comenti kuwa Hufanywa kwa njia ya Inspection yaani makadilio hili swali ni pamoja na lile la 3^x=9x huwa hakuna njia maalumu zaidi ya kupachika number na kuzijaribu kwenye equation.ASANTEN N POLEN KWA KUMALIZA ROUGH KUTAFUTA JIBU. Na pia kama bwana DASA unabisha toa njia niikosoe.

 

[Billie]

kwa kuongezea ni kuwa swali hili limetapakaa sana miongoni mwa walimu wa tuishen kama wakina MTATA,MWALAMI,HIDEN na kwingineko hamna solution maalumu japo kuwa majibu yapo.

 

[simplemind]

2^x = 4x can be rewritten as f(x)= 2^x - 4x = 0. Mathematically speaking f(x) is said to have two roots, moja ni x=4 na nyengine ni x=0.31 (2^0.31=1.24 na 4x0.31=1.24). Hii ndio jawabu kamili ie f(x) has two real roots.

 

[simplemind]

hilo swali linakuwa solved by graph:

kama hivi ifuatavyo, let y=2^x=4x then chora graph ya y=2^x, then, katika same axis chora graph ya y=4x, then pale graphs hizo zitakapo intersect ndiyo jibu lenyewe.

No. Chora graph ya y=2^x-4x jawabu ipo intersection ya graph na x-axis. Ie x=0.31 na x=4.

 

[Anheuser]

xlog 2-2log 2=log x
log 2 (x-2)=log x cancel log throughout
2(x-2)=x

Mtumee!

Huwezi ku cancel the logs throughout kwa sababu the function on the left is not log2(x-2) bali ni (log2)x-2

which are completely different functions.

Wrong!

 

[simplemind]

Nimepitia mabishano na solution zenu pamoja na thread ya huyu jamaa nikiwa na T.O wa advanced mathematics mwaka 2009 ametoka PUGU na kwa sasa yupo UDSm anasoma BCOM finance Tumecheka sana kutokana kuwa hili swali tumeshalifanyia kazi na huyu T.O ELINEEMA MOLA alipata 98% ya Pure math form six.
JIBU LA SWALI KUNA JAMAA MMOJA kapata ameandika comenti kuwa Hufanywa kwa njia ya Inspection yaani makadilio hili swali ni pamoja na lile la 3^x=9x huwa hakuna njia maalumu zaidi ya kupachika number na kuzijaribu kwenye equation.ASANTEN N POLEN KWA KUMALIZA ROUGH KUTAFUTA JIBU. Na pia kama bwana DASA unabisha toa njia niikosoe.

very true like most equation in science and engineering it has no exact solution and must be solved graphically or by numerical methods in conjuction with a computer mfano using newton method of succesive approximation. Swali limetoa changamoto na la kuelimisha.

 

[reandle]

Wale wataalam wa hesabu hebu nisaidieni namna ya kusolve hii haraka, nimejaribu njia zote nimeshindwa (Log... etc), sasa sijui ni uzee unanisumbua!!

2[SUP]x[/SUP]=4X

The answer is 4, sasa tatizo ni namna ya kuipata hiyo 4.
​

mkuu, ru serious au unafanya mzaha?

 

[Billie]

very true like most equation in science and engineering it has no exact solution and must be solved graphically or by numerical methods in conjuction with a computer mfano using newton method of succesive approximation. Swali limetoa changamoto na la kuelimisha.

Simplemind asante kuwa pamoja kwenye hili.