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香港!

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oh or is it 1.1^n?
i thought it was question 1.. different 1^n lol

anyway
y=1.1^n
ln y=n ln 1.1
n=(1\ln 1.1) ln y
dn\dy=(1\ln 1.1) (1\y)
dy\dn=flip it..
 
I

icycloud

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Dreamerish*~ said:
1.1n with respect to n.
Let y = 1.1^n

ln(y) = ln(1.1^n)
ln(y) = n * ln(1.1)
1/y * dy/dn = ln(1.1)

dy/dn = ln(1.1) * y
= ln(1.1) * 1.1^n #
 

Dreamerish*~

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香港! said:
0
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I'm afraid that is incorrect.

Oh, my bad. :eek: This is actually 2U. I feel like crap now.

Thanks Icycloud. Very smart. ;)
 

KFunk

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Dreamerish*~ said:
1.1n with respect to n.
Another way you could do it is using the property (a)<sup>n</sup> = e<sup>n.ln(a)</sup>, if you do this then:

y = 1.1<sup>n</sup> = e<sup>n.ln(1.1)</sup>

dy/dn = ln(1.1)e<sup>n.ln(1.1)</sup> = ln(1.1)(1.1<sup>n</sup>) or = ln(1.1)y
 

chin music

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y=1.1to the n
lny=ln(1.1 to the n)
lny=n*ln1.1 (plog rule or whatever other ppl call it)
n=1/ln1.1*(lny)
dn/dy=1/yln1.1
dy/dn = yln1.1
I reckon thats the best way of doing those
 

ioniser

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hey do you people know that way with the e function trick to answer the question

to answer the above ,where you make the equation in terms of e it was done at the 4 unit day at that uni thingie,its really usefull if your in any trouble but i cant remember it

(and know it wasn't the solution as above)
 
I

icycloud

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ioniser said:
hey do you people know that way with the e function trick to answer the question

to answer the above ,where you make the equation in terms of e it was done at the 4 unit day at that uni thingie,its really usefull if your in any trouble but i cant remember it

(and know it wasn't the solution as above)
Let x = e^y

ln(x) = ln(e^y)
= yln(e)
=y

Therefore, x = e^(ln(x))

Is that what you wanted?
 

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