exponential and logarithmic functions (1 Viewer)

[danno]

this topic is really horrible for me. can anyone supply me with some tips or anything?

(i basically need ANYTHING so whatever you know would be helpful)

thanks

 

[Estel]

eh
log rules are like arg rules.... oops that dont' help.
my tip is to go grab coroneos/fitz/cambridge and do an exercise on logs, and den do an exercise on expos.
if you don't know em, it's a good hr and a bit well spent.
den get a nice nite's sleep.

 

[bloodysunday]

what, like differentiation and integration of e's and ln's and so on? Give us an example of something that stumps you

 

[danno]

i know some stuff like how if you have e^x = 10 or something then ln10 = x.

apart from that i really dont know much.

also how do you integrate x^-1??

 

[smallcattle]

i know some stuff like how if you have e^x = 10 or something then ln10 = x.

apart from that i really dont know much.

also how do you integrate x^-1??

ln x + C ,,,,,,,,,,,,,

 

[danno]

ok heres a question. find the integral of log10x dx.

how do u integrate log10x? i know with e^2x the integral is (e^2x)/2. thats right isnt it?

 

[Doogsy]

jost remember the main rules:
ln(x)+ln(y)=ln(xy)
ln(x)-ln(y)=ln(x/y)
ln(x^y) = yln(x)
e^ln(x) = x
log<sub>2</sup>3= ln3/ln2
i think thats about it for logs.

 

[danno]

ln x + C ,,,,,,,,,,,,,

thanks. i really didnt pay attention to that stuff in class, so any little things like that would be really appreciated.

 

[Estel]

ur right

int[log(10)x]=lnt[lnx/ln10]=xlnx/ln10 - x/ln10 + C

 

[danno]

thanks doogsy.

how about integrating/differentiating log10x??

 

[~*HSC 4 life*~]

just know your rules, and remember to keep it simple..try to think to your self that e is just a number, like 2

might make things easier

 

[Estel]

d/dx (log(10)x) = d/dx (lnx/ln10) = 1/[xln10]... this is a q dey mite actually ask

 

[Doogsy]

d/dx (log(10)x) = d/dx (lnx/ln10) = 1/[xln10]... this is a q dey mite actually ask

yeah that one is trickey

 

[danno]

ur right

int[log(10)x]=lnt[lnx/ln10]=xlnx/ln10 - x/ln10 + C

ok so if i convert things into ln, i can differentiate and integrate them?

 

[bloodysunday]

general rule:

int.(a/bx+c) = a*ln(bx+c)+C

 

[Estel]

that is correct (to the person askin the q)

 

[danno]

so whats the basic rule for integrating lnx/lny?
and the basic one for differentiating?

 

[danno]

general rule:

int.(a/bx+c) = a*ln(bx+c)+C

lol your all too quick

do u have the differentiation one too?

 

[Estel]

to integrate lnx/lny ..... eh.... don't worry

If you're curious (and I always like curious mathy types)
y' = vu' + uv'
:. lnt(vu') = y - lnt(uv')
and set u' as 1 (u = x) and v as lnx
but be aware dat dis would be an interestin q 10 (developed of course)


dy/dx lnx/lny is simply 1/(xlny) since 1/lny is a constant.

 

[danno]

to integrate lnx/lny ..... eh.... don't worry

If you're curious (and I always like curios mathy types)
y' = vu' + uv'
:. lnt(vu') = y - lnt(uv')
and set u' as 1 (u = x) and v as lnx
but be aware dat dis would be an interestin q 10 (developed of course)


dy/dx lnx/lny is simply 1/(xlny) since 1/lny is a constant.

we wont be asked to integrate lnx/lny?

thanks for the dy/dx bit.