Could anyone explain what "ln" is and how to solve ln x + ln(2x + 1)=0?

Question

Could anyone explain what is "ln" and how to solve
ln x + ln(2x + 1)=0


Thank you very much!! :p

Answer 1

ln is simply the natural logarithm, which is the inverse of the exponential function with base e. It obeys the same properties as the any other logarithm, so in particular: ln x + ln y = ln (xy). So:

ln x + ln (2x+1) = 0
ln (x(2x+1))=0

Now exponentiate both sides:

x(2x+1)=e^0=1

Now solve this as a quadratic:

2x²+x=1
x²+x/2=1/2
x²+x/2+1/16=9/16
(x+1/4)²=9/16
x+1/4 = ±3/4
x=-1/4±3/4
x=-1 or x=1/2

However you remember that at the beginning you had:

ln x + ln (2x+1)

Which implies that ln x exists. Assuming you are not using complex logs (a safe assumption, considering that you don't even know what ln is), this means ln of a negative number does not exist. Therefore, x≠-1, so the unique solution is x=1/2

Answer 2

"ln" is called the "natural log". It's a logarithm whose base number is e, where e = 2.71828...

To solve a problem like ln(x) + ln(2x+1) = 0, just apply some of the rules with logs in general. First of all, there's log(a) + log(b) = log(ab). So applying this, you get ln (x*(2x+1)) = 0, or ln (2x^2 + x) = 0.

Second of all, there's the defintion of a log as it relates back to exponents: if you have Log[base b](a) = c, then a = b^c. So applying this to, we get (2x^2 + x) = e^0, so 2x^2 + x = 1. Now you just have to solve this quadratic for x. This becomes 2x^2 + x - 1 = 0, or (2x - 1)(x + 1) = 0. So x = 1/2 or -1. But you can't take the log of a negative number, so x=-1 won't solve the original equation. Therefore, the solution is x=1/2

(Also, Isabela asked what she might have been doing wrong. The rule to use is log(a)-log(b)=log(a/b). It looks like she had this switched around to log(a)/log(b) = log (a-b), which isn't necessarily true.)

Answer 3

"ln" is natural log, that is to say, log to the base e. e is a constant with the approximate value of 2.718281828459045...

Now to solve for x.

ln x + ln(2x + 1) = 0
ln{x(2x + 1)} = 0

Exponentiating we have

x(2x + 1) = e^0 = 1
2x² + x = 1
2x² + x - 1 = 0
(2x - 1)(x + 1) = 0
x = 1/2, -1

The solution x = -1 must be rejected because you cannot take the log of a negative number. Therefore

x = 1/2

Answer 4

ln means natural logarithm

ln x + ln(2x+1)= 0 can be express ln x(2x+1)=0 -property of logarithm

Answer 5

"ln" is the usual abbreviation for Napierian logarithms, or logarithm to the base "e", which is 2.71828... This particular equation may be solved by exponentiating both sides to get:
x(2x+1) = 1, and solving by the quadratic formula.

Answer 6

ln means the natural logarithm which is e to some power

Answer 7

ln means natural logarithm. As for solving it, sorry it's been a few years, I don't quite remember.

Answer 8

Ln is the natural log (logarithm) = log base e to the whatever follows the notation ln.
e=2.718281828459045

Note to ppl giving me "Thumbs Down"

Please leave an answer to this problem solved correctly for everyone to benefit from. or a reason why I am wrong.


Sorry I was wrong. Thanks.

Answer 9

ln is natural logarithm. e^lna is a itself .so take exponential on both sides of the equation. in this case it reduces to a linear equation.
P.S. the first 2 answers r wrong.

Answer 10

ln is logaritm
ln x + ln(2x + 1)=ln(x(2x+1))
ln1=0
ln(x(2x+1))=ln1
x(2x+1)=1
2x^2+x-1=0
x=-1
x=1/2