# Natural Log equation questionWatch

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#1
the quesstion is

Ln(x+3) + Ln(X-1)= 0

My thoguhts are if I do the inverse and put e in, then it cancels out to a simple equation, but that means the X's cancel out then? Also the answer has a Root 5 in it which is confusing.
0
3 years ago
#2
(Original post by SunDun111)
the quesstion is

Ln(x+3) + Ln(X-1)= 0

My thoguhts are if I do the inverse and put e in, then it cancels out to a simple equation, but that means the X's cancel out then? Also the answer has a Root 5 in it which is confusing. 0
3 years ago
#3
(Original post by SunDun111)
the quesstion is

Ln(x+3) + Ln(X-1)= 0

My thoguhts are if I do the inverse and put e in, then it cancels out to a simple equation, but that means the X's cancel out then? Also the answer has a Root 5 in it which is confusing.
It sounds like what you did was: but this isn't allowed.

The correct version is .

There's a difference.
0
#4
Thanks, just a quick question if you have something like 4x = Ln2
Do you write it as X = 1/4Ln2, cant you just write it as the entire Log divided by 4.
0
3 years ago
#5
Ln(x+3)= -ln(x-1)
ln(x+3) =ln(x-1)^-1
take e of both sides
x+3 = 1/x-1
solve from there
0
3 years ago
#6
(Original post by BrwnSugR1)
My initial thought is to just take the ln of both components and get "x+3+x-1=0" and solve for x from there.
Nope, that's wrong.

But since you say there is a root 5, the only way I found to get a root 5 in the answer is to do the following:

Ln(x+3)= -ln(x-1)
ln(x+3) =ln(x-1)^-1
This is fine, but overly lengthy.

take ln of both sides
No. You mean take the exponential of both sides.

x+3 = 1/x-1
solve from there and you will get a root 5 for x using the quadratic formula.
This is fine but overly lengthy.
0
3 years ago
#7
(Original post by SunDun111)
Thanks, just a quick question if you have something like 4x = Ln2
Do you write it as X = 1/4Ln2, cant you just write it as the entire Log divided by 4. . That's it, I'm not sure what question you're asking?
0
3 years ago
#8
(Original post by Zacken)
Nope, that's wrong.

This is fine, but overly lengthy.

No. You mean take the exponential of both sides.

This is fine but overly lengthy.
Yeah, you're right thanks!
0
#9
The way that the answerbook has it is that its written as 1/4Ln2 so i was just wondering what is the correct way to write it.
0
3 years ago
#10
(Original post by SunDun111)
The way that the answerbook has it is that its written as 1/4Ln2 so i was just wondering what is the correct way to write it.
The textbook means which is just as valid and equivalent to .
0
3 years ago
#11
(Original post by BrwnSugR1)
Yeah, you're right thanks!
Have a look at my reply to see how to do it! 0
3 years ago
#12
(Original post by SunDun111)
The way that the answerbook has it is that its written as 1/4Ln2 so i was just wondering what is the correct way to write it.
They both mean the exact same thing, just another way of writing it
0
#13
(Original post by KaylaB)
They both mean the exact same thing, just another way of writing it
Thanks was just being a bit dumb
0
3 years ago
#14
Where the brackets are expanded, I think the expansion should instead be x^2-x+3x-3=1 simplified to x^2+2x-4=0
1
3 years ago
#15
(Original post by BrwnSugR1)
Where the brackets are expanded, I think the expansion should instead be x^2-x+3x-3=1 simplified to x^2+2x-4=0
Yep, thank you!
0
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