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Natural Log equation question

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.

Reply 1

Zacken

Original post by SunDun111

Unparseable latex formula:

\displaystyle [br]\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.

(edited 8 years ago)

Reply 2

Zacken

Original post by SunDun111

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.

Reply 3

SunDun111

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.

Reply 4

BrwnSugR1

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

(edited 8 years ago)

Reply 5

Zacken

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.

Reply 6

Zacken

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?

Reply 7

BrwnSugR1

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!

Reply 8

SunDun111

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.

Reply 9

Zacken

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}

Reply 10

Zacken

Original post by BrwnSugR1

Yeah, you're right thanks!

Have a look at my reply to see how to do it! :smile:

Reply 11

KaylaB

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

Reply 12

SunDun111

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

Reply 13

BrwnSugR1

Original post by Zacken

Unparseable latex formula:

\displaystyle [br]\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