Question

why is log(-e+e^3) the same as 1+ log(e^2-1)

Answer 1

I assume the log here is the natural log

\[\log(-e+e^3)=\log(e(-1+e^2)=\log(e)+\log(-1+e^2)=1+\log(e^2-1)\]

Answer 2

Ohhhhhhhhhhhhhh i understand it now THANKS A MILLION @ ZARKON

Answer 3

No problem :)

Answer 4

@ zarkon can you also help me solve this
\[\int\limits_{1}^{e} (2x+e)/(x^2+ex)\]

Answer 5

let \[u=x^2+ex\]

\[du=(2x+e)dx\]

Answer 6

yes and what do you get as an answer?

Answer 7

Looks like it is

\[\left.\ln(x^2+ex)\right|_{1}^{e}\]

\[\ln(e^2+ee)-\ln(1+e)=\ln(2e^2)-\ln(1+e)\]

Answer 8

thank you very much!