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 uff how to prove this?
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Posted on 10-27-10 9:11 PM     Reply [Subscribe]
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guys how to prove this ? if you have any idea please help me. thanks.

Posted on 10-27-10 9:23 PM     [Snapshot: 10]     Reply [Subscribe]
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@ comingsoon,

- hope this helps ...

Proof for the Power Rule

loga xn nloga x


Step 1: 
Let m = loga x

Step 2: Write in exponent form 
x = am

Step 3: Raise both sides to the power of n
 = ( a)n

Step 4: Take log a of both sides and evaluate 
log a xn = log a amn
log a xn = mn log a a
log a xn = mn
log a xn = n loga x



Even a video on this 1: 

Last edited: 27-Oct-10 09:23 PM

Posted on 10-27-10 10:02 PM     [Snapshot: 61]     Reply [Subscribe]
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umm! thanks Black Panther .. ...... appreciated .

Posted on 10-28-10 10:34 AM     [Snapshot: 299]     Reply [Subscribe]
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Not sure if this is the way to solve , however trying is not bad.




logbMr=log(M1+M2+M3...+Mr)   //m1,m2,m3 =M ; 1,2,3 represents index  


M1, M2, M3 ..Mr represents same variable M there fore the next step is just arithmetic addition, i.e addition of r variable =rx





Posted on 10-28-10 2:55 PM     [Snapshot: 449]     Reply [Subscribe]
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@Black_Panther, Did you only copy paste the solution?
How would it be:
log a xn = log a amn
log a xn = mn log a a

Without actually prooving loga xn nloga x ? It is using the same thing which was actually asked to prove.
Posted on 10-28-10 3:45 PM     [Snapshot: 480]     Reply [Subscribe]
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- i knew someone would point that out ... sooner or later ... 
- i think the proof in the VDO does a better job ... 

Posted on 10-28-10 9:21 PM     [Snapshot: 584]     Reply [Subscribe]
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Last edited: 28-Oct-10 09:23 PM

Posted on 10-29-10 2:33 AM     [Snapshot: 662]     Reply [Subscribe]
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logbMr=log(M1+M2+M3...+Mr)   //m1,m2,m3 =M ; 1,2,3 represents index  

was supposed to be


logbMr=log(M1.M2.M3....Mr)   //m1,m2,m3 =M ; 1,2,3 represents index  




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