# how do i solve for x ?

ln x^2 + ln 2x+1 =0 thanks for your help

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Remember this natural log rule:

ln(a x b) = ln a + ln b

Therefore:

ln x^2 + ln 2x+1 = 0

equals:

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.

You get:

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... *

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

check:

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

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

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....

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.

1x1=1

1+1=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

1x+1x=2X

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