Question

help please how would you solve this

Answer 1

\[\ln(x-2)-\frac{ 1 }{ 2 }\ln(x-1)=-\frac{ 1 }{ 2 }\ln(y+1)+\frac{ 1 }{ 2 }\ln(y-1)\]

Answer 2

@sillybilly123

Answer 3

i already did the partial fraction already and thats what i got on both sides....

Answer 4

\(\ln(x-2)-\frac{ 1 }{ 2 }\ln(x-1) = \ln(x-2)-\ln(x-1)^{\frac{1}{2}} = \ln \dfrac{x-2}{(x-1)^{\frac{1}{2}}}\)

Answer 5

the other one?

Answer 6

\(-\frac{ 1 }{ 2 }\ln(y+1)+\frac{ 1 }{ 2 }\ln(y-1)\)


\(= -\frac{ 1 }{ 2 } \ln( \dfrac{y+1}{y-1})\)

Answer 7

really sorry Marcie, I may not be concentrating as well as I should on this.

Answer 8

oh no D: .. its due tomorrow morning e.e

Answer 9

ok

Answer 10

i cant seem to get the answer...
since the answer is

Answer 11

they want you to find the value of c

Answer 12

okay. so how do i do ...

Answer 13

hmm okayy so how do i do that bc i followed ur steps but i ended up with a different answer D:

Answer 14

this is a different thread, you present some algebra from some other thread, and then suggest that I have mislead you on something?!?!

you're not for real ?

Answer 15

im just llost how they got that answer

Answer 16

Marcie, I am about to be suspended for nothing. I will, pending suspension, try to help.

Answer 17

wait whaa D:

Answer 18

no no im not saying that u mislead me... i followed ur steps.. its just that i got lost till the end of the eq

Answer 19

it's fine. :)

Answer 20

no no im not saying that u mislead me... i followed ur steps.. its just that i got lost till the end of the equation..

Answer 21

ah, i remember typing that out

Answer 22

this is what you posted from the previous post... i just got to the partial fraction. im trying to figure out how they got to the answer. im very bad at logs

Answer 23

OK!!

Answer 24

logs are actually very important

Answer 25

well yeh... thats where im stucked but im not getting to the answer...

Answer 26

\(\log A + \log B = \log AB\)

Answer 27

mmk

Answer 28

so then

Answer 29

this was number 9 and this is what i did

Answer 30

OK

too tired tocheck anything but if YOU say that \(c = \ln C\) then you can add it into the logs on both sides and undo the Log thing. Make sense?

Answer 31

uhh

Answer 32

lets say that:


\(\ln Y = \ln X + C\)

And that: \(C = \ln C'\)

So:

\(\ln Y = \ln X + \ln C'\)

Or:

\(\ln Y = \ln C' X \)

Answer 33

what i mean is get that constant INSIDE the log

Answer 34

if is some constant so is log (C')

but, in the latter form, you can stuff it inside a log function

Answer 35

https://prnt.sc/i6lk0x

Answer 36

\[\frac{ 1 }{ 2 }(-\ln(y+1)+\ln(y-1))\]
\[\frac{1}{2} [ \ln(y-1)-\ln(y+1)] = \frac{1}{2} [\ln(\frac{y-1}{y+1})]\]

Answer 37

hmm ok

Answer 38

as he said on the other post it will be easier to multiply both sides by 2 to get rid of the fraction

Answer 39

yeah, "he" said that

Answer 40

he/she*

Answer 41

the trick is getting a constant C inside another constant log(C) so you can use log rules

Answer 42

Sorry i jumped ino the middle of this
please carry on

Answer 43

no you didn't.

Answer 44

Can you please carry on?

Answer 45

still there ??

Answer 46

https://prnt.sc/i6ltpk don't forget the constant when u integrate :=))

Answer 47

71 vs 69

there's only 1 winner there

Answer 48

🤔

Answer 49

good job..

Answer 50

TU, ma'am :)