log help！ (1 Viewer)

darkphoenix · Oct 8, 2011

Hey, i was doing log question and I cannot get the meaning of e^x
The question is to differentiate log10x and the answer is (log10e)/x
can some one explain why the e come out from no where. Thanks heaps

4025808 · Oct 8, 2011

darkphoenix said:
Hey, i was doing log question and I cannot get the meaning of e^x
The question is to differentiate log10x and the answer is (log10e)/x
can some one explain why the e come out from no where. Thanks heaps

log10(x) = loge(x)/loge(10) ------> change of base rule

now that 1/loge(10) is a constant, we leave it unchanged. then we differentiate loge(x) which gives 1/x

1/x * 1/loge(10) = 1/[xloge(10)]

btw the answer you provided is wrong, to confirm, go here -> http://www.wolframalpha.com/input/?i=differentiate+log+base+10+x

darkphoenix · Oct 8, 2011

OH, ok. Thanks, actually the full question(excel)is to differentiate with respect to x: (log10x)^3

and the solution is 1/x(3log10e)(log10x)^2

SpiralFlex · Oct 8, 2011

darkphoenix said:
OH, ok. Thanks, actually the full question(excel)is to differentiate with respect to x: (log10x)^3

and the solution is 1/x(3log10e)(log10x)^2

$\frac{\mathrm{d} }{\mathrm{d} x}(\log_{10}x)^3$

In this case you cannot fully utilise your logarithmic laws. Or can we? Change of base.

First of all I will clearly use the change of base law.

$\log_{10}x=\frac{\log_{e}x}{\log_{e}10}$

Now we can finally derive!

$=3(\log_{10}x)^2*\frac{1}{\log_{e}10}*\frac{1}{x}$

$=(\frac{3}{x}\log_{10}e) (\log_{10}x)^2$

darkphoenix · Oct 8, 2011

Sorry I just started to do log in this holiday, I am confused at change of base formula. log a(b)= logc(b) /logc(a) can c be any thing? or just e

math man · Oct 8, 2011

SpiralFlex said:
$\frac{\mathrm{d} }{\mathrm{d} x}(\log_{10}x)^3$

In this case you cannot fully utilise your logarithmic laws. Or can we? Change of base.

First of all I will clearly use the change of base law.

$\log_{10}x=\frac{\log_{e}x}{\log_{e}10}$

Now we can finally derive!

$=3(\log_{10}x)^2*\frac{1}{\log_{e}10}$

$=(3\log_{10}e) (\log_{10}x)^2$

answer should be $\frac{3}{xln10}(\log_{10}x)^{2}$

darkphoenix · Oct 8, 2011

So, confused, which one is right? all different answers

math man · Oct 8, 2011

i just saw the asnwer..to change my answer to the right answer you realise:

$\log_{e}10 = \frac{\log_{10}10}{\log_{10}e}$

this simplys to:

$\log_{e}10 = \frac{1}{\log_{10}e}$

and subbing this into my answer gives the desired result:

math man · Oct 8, 2011

fine full solution is:

$\frac{d}{dx}(\log_{10}x)^{3}=\frac{d}{dx}(\frac{\log_{e}x}{\log_{e}10})^{3}$ using change of base rule.

$\frac{d}{dx}(\frac{\log_{e}x}{\log_{e}10})^{3} = 3(\frac{1}{x\log_{e}10})(\frac{\log_{e}x}{\log_{e}10})^{2}$ using power rule.

Edit: full worked solution to obtain 1/x(3log10e)(log10x)^2 is:

from
$\frac{d}{dx}(\frac{\log_{e}x}{\log_{e}10})^{3} = 3(\frac{1}{x\log_{e}10})(\frac{\log_{e}x}{\log_{e}10})^{2}$

we realise that:

$(\frac{\log_{e}x}{\log_{e}10})= \log_{10}x$

and that

$\log_{e}10 = \frac{\log_{10}10}{\log_{10}e}$ by applying change of base rule for 10 now

subbing these into my first derivate becomes:

$\frac{d}{dx}(\frac{\log_{e}x}{\log_{e}10})^{3} = 3(\frac{1}{x(\frac{1}{\log_{10}e})})(\log_{10}x)^{2}$

and this becomes:

$\frac{d}{dx}(\frac{\log_{e}x}{\log_{e}10})^{3} = 3(\frac{\log_{10}e}{x})(\log_{10}x)^{2}$

which is the solution you gave us

nightweaver066 · Oct 8, 2011

$\frac{d}{dx} (log_{10}x)^3$

$\text{First, let's change the log in to something nicer using the change of base rule.}$

$log^3_{10}x = \frac{ln^3x}{ln^3 10}$

$\text{ Put this in and differentiate.}$

$\frac{d}{dx} (\frac{ln^3x}{ln^3 10})$
$= 3ln^2 x \times \frac{1}{x} \times \frac{1}{ln^3 10}$
$= \frac{3ln^2 x}{xln^3 10}$

D94 · Oct 8, 2011

math man said:
answer should be $\frac{3}{xln10}(\log_{10}x)^{2}$

This is one version which is correct.

Edit: and above is another correct version.

math man · Oct 8, 2011

darkphoenix said:
Sorry I just started to do log in this holiday, I am confused at change of base formula. log a(b)= logc(b) /logc(a) can c be any thing? or just e

and we can use the change of base for anything, but we only know how to differentiate base e so thats why we change it to e in these questions

SpiralFlex · Oct 8, 2011

Oops left out a part.

darkphoenix · Oct 8, 2011

math man said:
and we can use the change of base for anything, but we only know how to differentiate base e so thats why we change it to e in these questions

finally I understand, thank you and the others

log help！ (1 Viewer)

darkphoenix

Member

4025808

Well-Known Member

darkphoenix

Member

SpiralFlex

Well-Known Member

darkphoenix

Member

math man

Member

darkphoenix

Member

math man

Member

math man

Member

nightweaver066

Well-Known Member

D94

New Member

math man

Member

SpiralFlex

Well-Known Member

darkphoenix

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)