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
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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) (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 = = .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) (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
or
or
or
So the possible answers are or
However, plugging in 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
Answer by mananth(16946) (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
calculate b^2-4ac
we get:
x= {0.728968371242625, -3.72896837124262}
Ignore negative value
x = 0.729
ln(x+1) - ln(x -2) a
=
we have.
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 ..
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