Question

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M E D A L + F A N

Answer 1

@jtug6 lend a hand?

Answer 2

mhm 1 sec

Answer 3

Okay so do you know the inverse of ln?

Answer 4

uh I don't think so

Answer 5

Ok well basically ln and e are inverses of eachother where e raised to the ln power cancels it out as does ln of e so

\[e^{\ln}\] goes away and
\[\ln(e)\] goes away as well

Answer 6

so ln 4 + ln(2x) = 2 means you need to first subtract ln 4 from both sides because we want x entirely isolated even though its inside of ln

Answer 7

so what is that?

Answer 8

from addition property of log functions, maybe combine both on the left to begin

ln(4) + ln(3x) ----> ln(12x)

Answer 9

sorry I had went away for a bit

Answer 10

use base e and raise everything to powers

\[\huge e^{\ln(12x)} = e^2 \]

Answer 11

That works too or you can just subtract ln4 take e raise it then divide by 3 to find x.

Answer 12

left with 12x=e^2, some number

Answer 13

yeah, all works the same

Answer 14

mhmmmmm :D

Answer 15

in that case, why not make ln(3x) --- ln(3) + ln(x) ....to follow that custom.. hah idk

Answer 16

lol

Answer 17

lollll