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

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Remember, when you factor a polynomial, the roots are values of x that make the polynomial equal to zero, not the other way around. So you need to start with an equation, not just an expression, and then factor it.
  • #1
Coco12
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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?
 
Last edited:
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  • #2
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?
 
  • #3
Coco12 said:

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?
Coco12 said:
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.
Coco12 said:
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.
Coco12 said:
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.
 
  • #4
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?
 
Last edited:
  • #5
Coco12 said:
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!
Coco12 said:
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.
 
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  • #6
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.
 

1. How do I solve this logarithmic equation?

To solve this equation, we need to use the quadratic formula: x = (-b ± √(b^2-4ac)) / 2a. In this equation, a = 2, b = -7, and c = 3. Plugging these values in, we get x = (7 ± √(49-24)) / 4. Simplifying further, we get x = (7 ± √25) / 4. This results in two solutions: x = 2 or x = 1/2.

2. What is a logarithmic equation?

A logarithmic equation is an equation in which the variable appears as an exponent. In this case, we have 2x^2-7x+3 = 0, where x is the exponent of the base 2.

3. Why do we need to use the quadratic formula to solve this equation?

We need to use the quadratic formula because the equation is in the form of ax^2+bx+c = 0, where a, b, and c are constants. This type of equation cannot be solved using basic algebraic methods.

4. Are there any other methods to solve this equation?

Yes, there are other methods to solve this equation such as factoring, completing the square, or using the graphing calculator. However, the quadratic formula is the most efficient and reliable method.

5. How can I check if my solution is correct?

You can check your solution by plugging in the values of x into the original equation and seeing if it equals to 0. For example, if we plug in x = 2, we get 2(2)^2-7(2)+3 = 0, which is true. This confirms that x = 2 is a valid solution.

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