Question

My Achilles heel...

Answer 1

you could write 3 as \( e^{\ln3 } \)

Answer 2

I cant find that in my equation page..

Answer 3

what happends to x?

Answer 4

So how would u go on from there?

Answer 5

use
\[ \left(a^b\right)^c = a^{bc} \]

Answer 6

\[ \int 3^x \ dx = \int \left(e^{\ln3}\right)^x \ dx\\
= \int e^{\ln 3 \ x} \ dx \]

Answer 7

which is in the form
\[ \int e^{ax} \ dx \]

Answer 8

ok please continue, Ive had problem with these types, but ur explanation makes it easier.

Answer 9

can you integrate
\[ \int e^{ax} \ dx \]
?
this is a "standard" integral

Answer 10

Ive forgotten

Answer 11

what is the derivative of e^x ?

Answer 12

I have a problem with math, if I dont keep doing it, I fforget it..

Answer 13

this one is worth looking up

Answer 14

okey, I will go through that chapter again...can u guide me through this one?

Answer 15

\[ \frac{d}{dx} e^{a x}= e^{ax} \frac{d}{dx} ax = a\ e^{ax}\]

Answer 16

your problem is to "undo that"
in other words
\[ \int e^u \ du \]
where u = ax
\[ du = a \ dx\]

Answer 17

and
\[ \int e^u \ du = e^u \]

Answer 18

ok, I will have to go through this chapter again, I seem to be forgetting..also Im tired, thank you !:)

Answer 19

Trick:
to find \(\int 3^x dx\), we need find "something" whose derivative = \(3^x\), right?
let see, \((3^x)'= 3^x ln 3\) not \(3^x\), but ln3 is a constant, so, we just divide the original one by ln3. I meant
\((\dfrac{3^x}{ln3})' = (\dfrac{1}{ln3} *3^x)'= \dfrac{1}{\cancel{ln3}}3^x
*\cancel{ln3}=3^x\)
hence \(\int 3^x dx = \dfrac{3^x}{ln3}+C\)