SOLUTION: Solve the equation. Use a calculator to obtain decimal approximations, correct to two decimal places. ln (2x+3)+ln(x-6)-2lnx = 0 I know the answer is 10.68 but im just not su

Question 898387: Solve the equation. Use a calculator to obtain decimal approximations, correct to two decimal places.
ln (2x+3)+ln(x-6)-2lnx = 0
I know the answer is 10.68 but im just not sure how my teacher got it. Any help would be appreciated. Thank you.
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
ln (2x+3)+ln(x-6)-2lnx = 0
I know the answer is 10.68
-------------
Log Law:: log(A)+log(B) = log(AB)
Log Law:: log(A)-log(B) = log(A/B)
Log Law: log(A^n) = n*log(A)
----------------
Your problem::
ln[(2x+3)(x-6)/x^2] = 0
-----
(2x+3)(x-6)/x^2 = 1
---
2x^2 +3x-12x-18 = x^2
---
x^2 -9x - 18 = 0
---
Use the Quadratic Formula to get:
x = [9 +- sqrt(81-4*1*-18)]/2
x = [9 +- sqrt(153)]/2
x = [9 +- 12.37]/2
Positive solution::
x = 10.68
=============
Cheers,
Stan H.
--------------

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your equation is:

ln(2x+3) + ln(x-6) - 2*ln(x) = 0

since ln(2x+3) + ln(x-6) = ln((2x+3) * (x-6)), your equation becomes:

ln((2x+3)*(x-6)) - 2*ln(x) = 0

since 2*ln(x) = ln(x^2) and since ln((2x+3)*(x-6)) - ln(x^2) = ln((2x+3)*(x-6)/x^2), your equation becomes:

ln((2x+3)*(x-6)/x^2) = 0

this is true if and only if:

e^0 = (2x+3)*(x-6) / x^2

since e^0 = 1, your equation becomes:

1 = (2x+3)*(x-6) / x^2

multiply both sides of this equation by x^2 and simplify to get:

x^2 = 2x^2 - 12x + 3x - 18 = 2x^2 - 9x - 18

subtract x^2 from both sides of this equation to get:

0 = x^2 - 9x - 18

factor this equation using the quadratic formula to get:

x = 10.68465... or x = -1.68465...

since x has to be positive, the solution is x = 10.68465... which rounds to 10.68.

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