Find the solution if it exists.

[mathgeek69]

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 ?

 

[jedishrfu]

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.

 

[mathgeek69]

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 ?

 

[jedishrfu]

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.

 

[Ray Vickson]

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
x = \frac{e + 16 + 2\sqrt{5e+64}}{8 + \sqrt{5e+64}},
where e is the base of the natural logarithms: e = 2.7182818284590452354 ... .

 

[mathgeek69]

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.