SOLUTION: solve the natural logarithmic equation algebraically. Approximate the result to three decimal places. 1.) ln x + ln ( x + 3 ) = 1 2.) ln ( x + 1 ) - ln ( x - 2 ) = ln x^2 P

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: solve the natural logarithmic equation algebraically. Approximate the result to three decimal places. 1.) ln x + ln ( x + 3 ) = 1 2.) ln ( x + 1 ) - ln ( x - 2 ) = ln x^2 P      Log On


   

Question 668792: solve the natural logarithmic equation algebraically. Approximate the result to three decimal places.
1.) ln x + ln ( x + 3 ) = 1
2.) ln ( x + 1 ) - ln ( x - 2 ) = ln x^2
Please help me! I need it by today I have been trying I don't get it:(
Found 3 solutions by swincher4391, jim_thompson5910, mananth:
Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
ln(x) + ln(x+3) = 1
ln(x*(x+3) = 1
e^(ln(x*(x+3)) = e^1
x^2 + 3x = e
x^2 +3x -e = 0
By quadratic formula you get x = %281%2F2%29sqrt%289%2B4e%29+-3 = .729 as the only positive solution. Since ln(x) where x is negative is undefined.
ln(x+1) - ln(x-2) = ln(x^2)
ln((x+1)/(x-2)) = ln(x^2)
e^ln((x+1)/(x-2)) = e^ln(x^2)
(x+1)/(x-2) = x^2
x+1 = x^3 - 2x^2
x^3 -2x^2 -x -1 = 0
Find that x = 2.55 is the only real solution.



Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do #1 to get you started.

ln( x ) + ln ( x + 3 ) = 1

ln( x(x + 3) ) = 1

x(x + 3) = e^1

x(x + 3) = e

x^2 + 3x = e

x^2 + 3x - e = 0

x^2 + 3x - 2.71828 = 0 (note: e is roughly 2.71828...)

Now use the quadratic formula to solve for x. In this case, a = 1, b = 3, and c = 2.71828

x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%283%29%2B-sqrt%28%283%29%5E2-4%281%29%28-2.71828%29%29%29%2F%282%281%29%29

x+=+%28-3%2B-sqrt%289-%28-10.87312%29%29%29%2F%282%29

x+=+%28-3%2B-sqrt%289%2B10.87312%29%29%2F%282%29

x+=+%28-3%2B-sqrt%2819.87312%29%29%2F2

x+=+%28-3%2Bsqrt%2819.87312%29%29%2F2 or x+=+%28-3-sqrt%2819.87312%29%29%2F2

x+=+%28-3%2B4.4579278%29%2F2 or x+=+%28-3-4.4579278%29%2F2

x+=+0.7289639 or x+=+-3.7289639

So the possible answers are x+=+0.7289639 or x+=+-3.7289639

However, plugging in x+=+-3.7289639 yields a negative argument in the log, which is not allowed. So we must toss out this x value.

So the only solution (to 3 decimal places) is x+=+0.729

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
ln x + ln ( x + 3 ) = 1
lnx + ln (x+3) = 1
ln x(x+3) = 1
x(x+3) = e^1
x^2+3x = e^1
x^2+3x-e = 0
x^2+3x-2.7183 = 0
compare with ax^2+bx+c
a=1,b=3,c=-2.7183
discriminant
x=%28-b%2B-sqrt%28b%5E2-4%2Aa%2Ac%29%29%2F%282%2Aa%29

calculate b^2-4ac
we get:
x= {0.728968371242625, -3.72896837124262}
Ignore negative value
x = 0.729

ln(x+1) - ln(x -2) a
=+ln+%28%28x%2B1%29%2F+%28x-2%29%29
we have.
ln%28%28x%2B1%29%2F+%28x-2%29%29+=+ln%28x%5E2%29
x+1
____ = x^2
x -2
cross multiply
x^2 (x -2) = x+1
x^3 -2x^2-x -1 = 0
The only way I know is to graph it ..