maheshmeghwal9
  • maheshmeghwal9
Plz help:)
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
hey mahesh
maheshmeghwal9
  • maheshmeghwal9
1 Attachment
maheshmeghwal9
  • maheshmeghwal9
why b>1? Why not b>0?

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maheshmeghwal9
  • maheshmeghwal9
in the attachment.
anonymous
  • anonymous
holymoly that's one big pic.
maheshmeghwal9
  • maheshmeghwal9
should I make it small?
anonymous
  • anonymous
sawrite....
maheshmeghwal9
  • maheshmeghwal9
1 Attachment
anonymous
  • anonymous
b>1 because the negatitve exponents will take care of values for b between 0 and 1... that make sense?
anonymous
  • anonymous
so it is defined for b>1
maheshmeghwal9
  • maheshmeghwal9
didn't understand ya main concept:(
anonymous
  • anonymous
let's do this then...
anonymous
  • anonymous
|dw:1337931693406:dw|
anonymous
  • anonymous
do you see why b doesn't have to be defined between 0 and 1?
maheshmeghwal9
  • maheshmeghwal9
but what is the problem for 0
anonymous
  • anonymous
there's no problem... with 0
maheshmeghwal9
  • maheshmeghwal9
k! did it have no need due to log?
anonymous
  • anonymous
notice the base of logs CAN have a base between 0 and 1 also but could be transformed and written with a base bigger than 1
anonymous
  • anonymous
is there a reason why this was bumped?
maheshmeghwal9
  • maheshmeghwal9
Hi @experimentX I couldn't understand the my basic questiom.Plz help:)
experimentX
  • experimentX
I don't understand the reason why b should be greater than 1, but b should always be greater than zero ... for that expression to be defined
maheshmeghwal9
  • maheshmeghwal9
ya I also think this:)
experimentX
  • experimentX
\[ \log_b x = \frac{\log_k x}{\log_k b}\] Also between 0-1 \( \log_k b < 1\)
maheshmeghwal9
  • maheshmeghwal9
ok!
maheshmeghwal9
  • maheshmeghwal9
I had this problem while studying Logarithmic Differentiation. I think there is some connection between b>1 & Logarithmic Differentiation.
experimentX
  • experimentX
\[ \frac{d}{dx} (\log_b x) = \frac{1}{x \ln b}\] At b=1, \( \ln b =0\) is zero ... so it is undefined, so is the above expression, but for (0,1) the expression is defined http://www.wolframalpha.com/input/?i=plot+y+%3D+d%2Fdx+log[0.5%2C+x]+from+0+to+10 http://www.wolframalpha.com/input/?i=plot+y+%3D+log[0.5%2C+x]+from+0+to+10
maheshmeghwal9
  • maheshmeghwal9
So indirectly do u want to say that b>1 is right there! Please tell:)
experimentX
  • experimentX
No ... i think i got it wrong it somewhere check this out http://www.wolframalpha.com/input/?i=plot+y+%3D+log[x%2C+2]+from+0+to+10
maheshmeghwal9
  • maheshmeghwal9
k! thanx alot:)
experimentX
  • experimentX
wait ...
maheshmeghwal9
  • maheshmeghwal9
k!
experimentX
  • experimentX
it seems i was right ... but somewhat not accurate http://www.wolframalpha.com/input/?i=plot+y+%3D+log[x%2C+5]+from+0+to+2
maheshmeghwal9
  • maheshmeghwal9
K! I understood that.I also think u r right.
experimentX
  • experimentX
at b=0, \( \log_b x\) is zero ... and \( dy/dx\) is indeterminate
maheshmeghwal9
  • maheshmeghwal9
ok! thanx.
maheshmeghwal9
  • maheshmeghwal9
ow I have some clues. I will think on that point more. Thanx for clues:)
maheshmeghwal9
  • maheshmeghwal9
ow=now
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=plot+y+%3D+d%2Fdx%28log[b%2C+x]%29+from+b%3D0+to+2 check this out
maheshmeghwal9
  • maheshmeghwal9
Hmm! I checked that:)
experimentX
  • experimentX
b cannot be 1 and less than zero ... that's all
maheshmeghwal9
  • maheshmeghwal9
k! but can be (0,1)
experimentX
  • experimentX
i guessed so ...
maheshmeghwal9
  • maheshmeghwal9
so we again came there; where was our doubt:)
maheshmeghwal9
  • maheshmeghwal9
this is a very confusing state; isn't it?
experimentX
  • experimentX
well yes!!
maheshmeghwal9
  • maheshmeghwal9
What should we do now? by the way I have asked this to my sir also online, but they will receive question & give answer after almost a week!!!
experimentX
  • experimentX
what was the original question??
maheshmeghwal9
  • maheshmeghwal9
Pardon me what do u want?; that is a video cutted part. I had this problem while studying video:)
experimentX
  • experimentX
Oh ... i thought this is a part of the problem where you got stuck!!
maheshmeghwal9
  • maheshmeghwal9
so is there anyone in OS who can solve our problem:)
experimentX
  • experimentX
I don't know ...
maheshmeghwal9
  • maheshmeghwal9
k! np:) Thanx for ur co-operation. When my problem is solved, i'll tell u too.
experimentX
  • experimentX
k

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