i desperately need help with these two problems!!

[jenzy569]

i understand all of them except these two problems:


-exponential function = y = e^x
-logarithmic function = y = logx



Plot the graph of the above equations.

 

[Aladdin]

i understand all of them except these two problems:


-exponential function = y = ex
-logarithmic function = y = logx



Plot the graph of the above equations.

F(x)=y=log(x)

Domain : ]0,+infinity[

Limits : x tends to 0 -- > y=log(0)= -infinity , x=0 is a verticle asymptote

x tends to +infinity --> y=log(+infinity)= + infinity

\(\displaystyle \frac{log(x)}{x}\)

\(\displaystyle \frac{\infty}{\infty}\)

---Hospitals Rule---

\(\displaystyle \frac{log(x)}{x}\)tends to zero as x tends to + infinity.

So Asymptotic direction parallel to x'x .

 

[red and white kop!]

logx is nothing: it doesnt have a base
but since it says 'logarithmic function' im suspecting that this is lnx, i.e. log(base e)x
lnx is the reflection of e^x in y=x
then you can figure out aladdin's extremely simple explanation to graph e^x and get lnx from there
:twisted:

 

[Aladdin]

logx is nothing: it doesnt have a base
but since it says 'logarithmic function' im suspecting that this is lnx, i.e. log(base e)x
lnx is the reflection of e^x in y=x
then you can figure out aladdin's extremely simple explanation to graph e^x and get lnx from there
:twisted:

Yes the inverse of exponential is logarithms, This is how we study any function.

 

[stapel]

-exponential function = y = e^x
-logarithmic function = y = logx
Plot the graph of the above equations.

I have a feeling the discussion of calculus topics may not have proved helpful in answering your algebra question. Sorry!

To learn how to do the graphing, try here and here. :wink:

Note: Once you get to post-calculus studies, you may encounter contexts in which "log(x)" refers to the base-e (that is, the "natural") logarithm. Usually, though, it refers to the base-10 (that is, the "common") log. The graph will depend upon the base so, if you're not sure, ask your instructor for clarification.

 

[jenzy569]

thank you all for your help!!