how do i solve for x ?
ln x^2 + ln 2x+1 =0 thanks for your help
- βread⊆ℜumbs™Lv 51 decade agoFavorite Answer
Remember this natural log rule:
ln(a x b) = ln a + ln b
ln x^2 + ln 2x+1 = 0
ln x^2(2x+1) = 0
Now take 'e' of both sides:
e^ln x^2(2x+1) = e^0
which gives you:
x^2(2x+1) = 1
Multiply out the left side. Then move '1' to the left side.
2x^3 + x^2 - 1 = 0
From here, I used a tweaked version of Cardano's Formula and got this:
x = 0.5897.
Hope this helps!Source(s): My fault notthejake....got sloppy with my own rules haha. *The tweaked version of Cardano's Formula : http://www.math.vanderbilt.edu/~schectex/courses/c... *
- notthejakeLv 71 decade ago
remembering that ln a + ln b = ln (ab), this becomes:
ln(x^2 * (2x + 1)) = 0
converting to exponential form:
e^0 = 2x^3 + x^2
but e^0 = 1, so 2x^3 + x^2 - 1 = 0
graphically, the only real solution is x = .657
ln (.657)^2 = -.8401
ln(2*.657 + 1) = ln 2.314 = .839
with rounding errors, -.840 + .839 = 0
Edit: careful breadcrumbs: e^0 = 1, not 0
- Jerome JLv 71 decade ago
ln x^2 + ln 2x + 1 = 0
2 ln x + ln 2 + ln x + 1 = 0
2 ln x + ln x = -1 - ln 2
3 ln x = -1 - ln 2
ln x = (-1 - ln 2) / 3
x = e^[(-1 - ln 2) / 3]
x = 0.5687
- fred osimLv 41 decade ago
ln x^2 + ln 2x+1=0, or 2lnx+ln2+lnx+1=0, or
3lnx=-1-ln2=-1.693 or lnx=-.564...or x=.569....done
check, ln (.569)^2+ln 2(.569)+1=0...or
-1.128+.1293+1=0...which check out...
so, ans is x=.569....
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- 1 decade ago
i dont know what the in is. but if the problem is x^2+2x+1=0 then all you have to do is simple factoring. so what you need to do here is find what multiples to c, in this case 1 and adds to b, in this case 2.
so the way to factor your problem would be (x+1)(x+1)
if you want to make sure this is correct, use foil, in other words distribute.
x times x=x^2
x times 1=1x
1 times x=1x
1 times 1=1
once you have your factor, (x+1)(x+1)
then you would need to set each factors equal to zero. x+1=0
thus x = -1
your answer is x equals negative one