Question

ln e^x+4y ??? im confused

Answer 1

why ?

Answer 2

because i forgot how to do it

Answer 3

what exactly are you trying to do ?

Answer 4

simplify

Answer 5

Ohkay \[\large \ln e^{x+4y}\]

like this ?

Answer 6

yes

Answer 7

use below log property :
\[\ln (a^b) = b\times \ln(a)\]

Answer 8

\[\large \ln (e^{x+4y}) = (x+4y) \times \ln (e)\]

Answer 9

what do i do ? idk how to begin

Answer 10

ln(e) = 1

Answer 11

so it means that ln cancels e

Answer 12

so is just the x+4y?

Answer 13

so you end up with just x+4y

Answer 14

\[\large \ln (e^{x+4y}) = (x+4y) \times \ln (e) = (x+4y)\times 1 = x+4y\]

yes thats the end of simplification.
There is a nice way to make sense of these identties by using the fact that \(\ln (x)\) and \(e^x\) are inverses of each other. Are you good with inverse functions and graphs ?

Answer 15

no :(

Answer 16

\[\ln(e ^{x+4y})=(x+4y) \]

Answer 17

yeah what @ganeshie8 says graph ln( and e^x so you can see it

Answer 18

its easy :