Find the solution if it exists.

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Homework Help Overview

The discussion revolves around solving the equation ln(5x+6) + ln(x-2) = 1, which involves logarithmic properties and quadratic equations. Participants are exploring the steps taken to manipulate the equation and are questioning the accuracy of their algebraic expansions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of logarithmic identities and the quadratic formula. There are attempts to clarify the algebraic manipulation of terms and the implications of errors in expansion. Questions arise regarding the precision of the solution and the correctness of the derived expressions.

Discussion Status

Some participants have provided feedback on the algebraic steps, indicating where corrections are needed. There is acknowledgment of typos affecting the clarity of the solution. The conversation reflects an ongoing exploration of the problem without reaching a definitive conclusion.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of guidance provided. There is a focus on ensuring that the mathematical reasoning is sound while navigating through potential misunderstandings.

mathgeek69
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1. ln(5x+6)+ ln(x-2)=1
2. Homework Equations : log(xy)= log(x)+ log(y)
3. The Attempt at a Solution :

ln(5x+6)+ ln(x-2)=1
ln((5x+6)(x-2))=1
ln(5(x^2)-14x-12)=1
e^(ln(5(x^2)-14x-12)) = e^1
5(x^2)-4x-12=e
5(x^2)-4x-(e-12)=0

use quadratic formula to find x and I get x=2.1617

I don't think I am doing this right. Do you see where my concepts went wrong ?
 
Last edited:
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Your logic looks good, but the quadratic expansion should be 5x^2 + 6x - 10x - 12 = e

Also when you brought e to the lefthand side that term should be - (12 + e)

Giving 5x^2 - 4x - (12+e) = 0

Now you can use the quadratic formula to find both x values.
 
jedishrfu said:
Your logic looks good, but the quadratic expansion should be 5x^2 + 6x - 10x - 12 = e


Sorry that's what I meant. It was a typo. So my x is a long decimal answer. Is there any way of finding a preciser and a more clear answer ?
 
mathgeek69 said:
Sorry that's what I meant. It was a typo. So my x is a long decimal answer. Is there any way of finding a preciser and a more clear answer ?

Well if you redefine e to be zero and change all of mathematics then maybe but as it stands now nope.
 
mathgeek69 said:
1. ln(5x+6)+ ln(x-2)=1



2. Homework Equations : log(xy)= log(x)+ log(y)



3. The Attempt at a Solution :

ln(5x+6)+ ln(x-2)=1
ln((5x+6)(x-2))=1
ln(5(x^2)-14x-12)=1
e^(ln(5(x^2)-14x-12)) = e^1
5(x^2)-4x-12=e
5(x^2)-4x-(e-12)=0

use quadratic formula to find x and I get x=2.1617

I don't think I am doing this right. Do you see where my concepts went wrong ?

Your 3rd and 4th lines are wrong, your 5th line is right, your 6th line is wrong. Magically, you still manage to get the correct answer! You really do need to be more careful. Mistakes like than on an exam can be very costly.

BTW: an "exact" answer would be
[tex]x = \frac{e + 16 + 2\sqrt{5e+64}}{8 + \sqrt{5e+64}},[/tex]
where e is the base of the natural logarithms: e = 2.7182818284590452354 ... .
 
Last edited:
Ray Vickson said:
Your 3rd and 4th lines are wrong, but your 5th and 6th lines are right---why the switch? Your final answer is right.

sorry it was a typo. instead of 14x i meant to write 4x...

thanks.
 

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