SunDun111
Badges: 2
Rep:
?
#1
Report Thread starter 3 years ago
#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
reply
Zacken
Badges: 22
Rep:
?
#2
Report 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.
\displaystyle 

\begin{align*}\ln(x+3) + \ln(x-1) = 0 &\iff \ln(x+3)(x-1) = 0 \\ &\iff (x+3)(x-1) = e^0 = 1 \\ &\iff x^2 -x + 3x - 3 = 1 \\ & \iff x^2 +2x -4 = 0 \\ &\iff \cdots\end{align*}

Solve this quadratic.
0
reply
Zacken
Badges: 22
Rep:
?
#3
Report 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:

\displaystyle \ln(x+3) + \ln (x-1) = 0 \Rightarrow e^{\ln (x+3)} + e^{\ln (x-1)} = e^0 but this isn't allowed.

The correct version is \displaystyle \ln(x+3) + \ln (x-1) = 0 \Rightarrow e^{\ln(x+3) + \ln (x-1)} = e^0.

There's a difference.
0
reply
SunDun111
Badges: 2
Rep:
?
#4
Report Thread starter 3 years ago
#4
(Original post by Zacken)
It sounds like what you did was:

\displaystyle \ln(x+3) + \ln (x-1) = 0 \Rightarrow e^{\ln (x+3)} + e^{\ln (x-1)} = e^0 but this isn't allowed.

The correct version is \displaystyle \ln(x+3) + \ln (x-1) = 0 \Rightarrow e^{\ln(x+3) + \ln (x-1)} = e^0.

There's a difference.
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
reply
BrwnSugR1
Badges: 1
Rep:
?
#5
Report 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
reply
Zacken
Badges: 22
Rep:
?
#6
Report 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
reply
Zacken
Badges: 22
Rep:
?
#7
Report 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.
4x = \ln 2 \iff x = \frac{\ln 2}{4}. That's it, I'm not sure what question you're asking?
0
reply
BrwnSugR1
Badges: 1
Rep:
?
#8
Report 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
reply
SunDun111
Badges: 2
Rep:
?
#9
Report Thread starter 3 years ago
#9
(Original post by Zacken)
4x = \ln 2 \iff x = \frac{\ln 2}{4}. That's it, I'm not sure what question you're asking?
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
reply
Zacken
Badges: 22
Rep:
?
#10
Report 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 x = \frac{1}{4}\ln 2 which is just as valid and equivalent to x = \frac{\ln 2}{4}.
0
reply
Zacken
Badges: 22
Rep:
?
#11
Report 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
reply
KaylaB
Badges: 20
Rep:
?
#12
Report 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
reply
SunDun111
Badges: 2
Rep:
?
#13
Report Thread starter 3 years ago
#13
(Original post by Zacken)
The textbook means x = \frac{1}{4}\ln 2 which is just as valid and equivalent to x = \frac{\ln 2}{4}.
(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
reply
BrwnSugR1
Badges: 1
Rep:
?
#14
Report 3 years ago
#14
(Original post by Zacken)
\displaystyle 

\begin{align*}\ln(x+3) + \ln(x-1) = 0 &\iff \ln(x+3)(x-1) = 0 \\ &\iff (x+3)(x-1) = e^0 = 1 \\ &\iff x^2 -x + 3x - 1 = 1 \\ & \iff x^2 +2x -2 = 0 \\ &\iff \cdots\end{align*}

Solve this quadratic.
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
reply
Zacken
Badges: 22
Rep:
?
#15
Report 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
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • The University of Law
    Solicitor Series: Assessing Trainee Skills – LPC, GDL and MA Law - Guildford campus Postgraduate
    Wed, 29 Jan '20
  • Nottingham Trent University
    Postgraduate Open Day Postgraduate
    Wed, 29 Jan '20
  • University of Groningen
    Undergraduate Open Day Undergraduate
    Fri, 31 Jan '20

Why do you want to do a masters?

Great for my career (8)
32%
I really love the subject (10)
40%
I don't know what else to do (5)
20%
I can't get a job (1)
4%
My parents want me to (1)
4%
I don't know... I just do (0)
0%

Watched Threads

View All