Question

how do i simplify 3e^loge 2 ?

Answer 1

\[\huge 3e^{\log_e 2}\]Is this what it looks like?

Answer 2

yes :)

Answer 3

The exponential and logarithmic function are `inverses` of one another, so they essentially "undo" each other.

\[\large f\left[f^{-1}(x)\right]=x\]

\[\huge e^{(\log_e x)}=x\]

Answer 4

So what does this give us?\[\huge e^{(\log_e 2)}=?\]

Answer 5

is it 2?

Answer 6

Yessss it is :)

And remember we had that big ugly thing being multiplied by 3.

So now our problem becomes,\[\huge 3\left(e^{\log_e 2}\right) \qquad=\qquad 3\left(2\right)\]

Answer 7

ohhhhh i seee
thats easy :D

Answer 8

hang on ive got another question

Answer 9

This idea of inverses is a little tricky.
It might take a few more problems before it really sticks in your brain :)

Answer 10

hahahahaa truuueee
how do i simplifyy -2e^-loge2?

Answer 11

i dont get how -loge2 is equal to 1/2 :(

Answer 12

You'll often see the log of base e written as the "natural log".
Let's do that, it'll make it a little easier to read without the e subscript.

\[\huge -2e^{-\log_e2} \qquad=\qquad -2e^{-\ln2}\]

Answer 13

Have you seen that before? :O If not it'll be confusing.

Answer 14

i think this is sort of the first timee
and i when i just saw that i just went blankk

Answer 15

Ok fine fine c: we'll stick with the log notation lol

Answer 16

bahahha okay

Answer 17

This problem is a little trickier than the last.
We're not allowed to use that same trick we did in the last problem, at least not yet.

The reason is because the negative sign is in the way.
We need to recall a rule of logarithms.\[\large b\cdot \log(a) \qquad=\qquad \log(a^b)\]

So let's apply this rule to our exponent,\[\large -1\cdot \log(2) \qquad=\qquad ?\]

Answer 18

ohhhh
so its 2^-1

Answer 19

rightt
and thats 1/2

Answer 20

ohhh i get it

Answer 21

Yesss good good.

Answer 22

thanks for helpinggg :)

Answer 23

no probs \c:/