# Thread: why is e^ log x = x

1. i understand the 2nd part, but why is the derivative 1/ x log e (e^ log x)?

by chain rule, i get 1/x (e^ log x)...becuase u keep the e^u and multiply by the derivative of u, which in this case is 1/x

2. Originally Posted by twostep08
i understand the 2nd part, but why is the derivative 1/ x log e (e^ log x)?

by chain rule, i get 1/x (e^ log x)...becuase u keep the e^u and multiply by the derivative of u, which in this case is 1/x
i dont know how that ended up on top, i meant to reply

3. Originally Posted by Chris L T521
Its also good to know that $\displaystyle \log e^x = e^{\log x}$

It turns out that if you differentiate $\displaystyle e^{\log x}$, you get $\displaystyle \frac{1}{x\log e}e^{\log x}$, but that is the same as $\displaystyle \frac{1}{\log e^x}e^{\log x}$. If we can assume the equality I mentioned, then we see that we end up with $\displaystyle \left[e^{\log x}\right]^{\prime}=1$.

Does this make sense?
i understand the 2nd part, but why is the derivative 1/ x log e (e^ log x)?

by chain rule, i get 1/x (e^ log x)...becuase u keep the e^u and multiply by the derivative of u, which in this case is 1/x

4. Originally Posted by twostep08
i understand the 2nd part, but why is the derivative 1/ x log e (e^ log x)?

by chain rule, i get 1/x (e^ log x)...becuase u keep the e^u and multiply by the derivative of u, which in this case is 1/x
Recall that $\displaystyle \left(\log_a x\right)^{\prime}=\frac{1}{x\ln a}$.

In the case of $\displaystyle \log x$ (which I assume is in base $\displaystyle e$), we see the derivative is $\displaystyle \frac{1}{x\log e}=\frac{1}{x}$. But for the sake of this problem, we're interested in the derivative of the form $\displaystyle \frac{1}{x\log e}=\frac{1}{\log e^x}$.

Does this clarify things?

5. Originally Posted by Chris L T521
Recall that $\displaystyle \left(\log_a x\right)^{\prime}=\frac{1}{x\ln a}$.

In the case of $\displaystyle \log x$ (which I assume is in base $\displaystyle e$), we see the derivative is $\displaystyle \frac{1}{x\log e}=\frac{1}{x}$. But for the sake of this problem, we're interested in the derivative of the form $\displaystyle \frac{1}{x\log e}=\frac{1}{\log e^x}$.

Does this clarify things?

oh ok that makes sense..thanks...the ln you threw in there initially confused me... in HS, we used ln alot, but now my college professor claims we were misguided by ln, and that log should be used in its place..

i dont know if that makes sense...anyways i digress,,, thanks for all the help- i got it now

6. ## why is e^ log x = x

i know the derivative e^ logx = 1
but i need to show that in my work somehow...by chain rule, i got 1/x e^logx
thats the exact same problem i started with, because if i can do something to show that e^logx =x then i get x/x which is 1

also, i got another similar log question that i asked a few mins. ago, that i accidently replied to myself- and thereford im thinking people are disregarding it as answered, so id appreciate it if someone could give that a look if its still unanswered

any help is very appreciated!

7. Originally Posted by twostep08
i know the derivative e^ logx = 1
but i need to show that in my work somehow...by chain rule, i got 1/x e^logx
thats the exact same problem i started with, because if i can do something to show that e^logx =x then i get x/x which is 1

also, i got another similar log question that i asked a few mins. ago, that i accidently replied to myself- and thereford im thinking people are disregarding it as answered, so id appreciate it if someone could give that a look if its still unanswered

any help is very appreciated!
Its also good to know that $\displaystyle \log e^x = e^{\log x}$

It turns out that if you differentiate $\displaystyle e^{\log x}$, you get $\displaystyle \frac{1}{x\log e}e^{\log x}$, but that is the same as $\displaystyle \frac{1}{\log e^x}e^{\log x}$. If we can assume the equality I mentioned, then we see that we end up with $\displaystyle \left[e^{\log x}\right]^{\prime}=1$.

Does this make sense?