Solving Logarithms: Find x in 2x^2-7x+3 = 0

[Coco12]

Homework Statement

Log(2x-3)=log(4x-3)-logx

Homework Equations

Logcl =logcr
L=r

The Attempt at a Solution

Quotient law of logs

So I simplfied:
Log(2x-3)=log(4x-3)/x

Since same base:
2x-3=4x-3/x

2x^2-7x+3=0
(X-1)(x-6)

The ans is supposed to be 3. I don't know what I'm doing wrong?

 

[ar1994]

Have you double checked your algebra? After you've removed the logarithms and come up with the algebraic equation, are you sure you manipulated it into a polynomial correctly?

 

[Mark44]

Homework Statement

Log(2x+5)=log(4x-3)-logx

Homework Equations

Logcl =logcr
L=r

The Attempt at a Solution

Quotient law of logs

So I simplfied:
Log(2x+3)=log(4x-3)/x

First it was log(2x + 5), but now it's log(2x + 3). Which one is it?

Since same base:
2x+3=4x-3/x

2x^2-7x+3
(X-1)(x-6)

The ans is supposed to be 3. I don't know what I'm doing wrong?

For the equation you posted, 3 is not a solution. Here's the check:
If x = 3,
log(11) =? log(9) - log(3)
log(11) =? log(9/3) = log(3)
Since log(11) ≠ log(3), 3 is not a solution.

In your work you need a few more parentheses.

So I simplfied:
Log(2x+3)=log(4x-3)/x

Since same base:
2x+3=4x-3/x

In the two equations above, log(4x-3)/x should be written as log[(4x - 3)/x], and 4x -3/x is not the same as (4x - 3)/x.

2x^2-7x+3
(X-1)(x-6)

In what you have above, you started with an equation, but lost the = 0, so you're not working with equations any more.


Have you copied the equation correctly? If so, did you look at the right solution? There's also the possibility that the answer in your book is wrong.

 

[Coco12]

Ya it's supposed to be 2x-3. That is what the book says.. That the ans is 3.

I think 3 is right. Because when I plugged in into the equation, I got it..

However how did they get that?

 

[Mark44]

Ya it's supposed to be 2x-3.

Which is different from either of the two expressions you had before. You need to be more careful!

That is what the book says.. That the ans is 3.
So besides missing the brackets, is the solution correct?

No.
Just before you did the factoring, you had 2x² - 7x + 3, which is correct as far as it goes.

It should be 2x² - 7x + 3 = 0.

Try again on the factorization.

You will also need to consider the domain for each individual log expression.
What are the restrictions on x so that log(2x - 3) is defined?
What are the restrictions on x so that log(4x - 3) is defined?
What are the restrictions on x so that log(x) is defined?

Your solution has to be in an interval for what all three log expressions are defined.

 

[ar1994]

Even if it is supposed to be 2x-3, you're factorization is still incorrect. To see why, simply try plugging in the roots you found back into the logarithmic equation, as Mark44 suggested, or even in the polynomial from which you obtained the roots. You will find that they do not satisfy the equation.