Is anything under 1 a ln?

[Jason76]

example:

\(\displaystyle \dfrac{1}{x} = \ln x\)

\(\displaystyle \dfrac{1}{4x - 3} = \ln (4x - 3)\)

 

[lookagain]

example:

\(\displaystyle \dfrac{1}{x} = \ln x \ \ \ \ \)False!

\(\displaystyle \dfrac{1}{4x - 3} = \ln (4x - 3) \ \ \ \ \)False!

Neither of those equations are true. They are not equalities.

 

[stapel]

example:

\(\displaystyle \dfrac{1}{x} = \ln x\)

\(\displaystyle \dfrac{1}{4x - 3} = \ln (4x - 3)\)

Are you, by any chance, asking something related to integration?

 

[lookagain]

example:

\(\displaystyle \dfrac{1}{x} = \ln x \ \ \ \ \ \) Correction: \(\displaystyle \ \ \int \dfrac{1}{x}dx \ = \ ln|x| \ + \ C\)

\(\displaystyle \ \dfrac{1}{4x - 3} = \ln (4x - 3)\)

Also, this is a correction: \(\displaystyle \ \ \int \dfrac{1}{x - 3} dx\ = \ \dfrac{ln|4x - 3|}{4} \ + \ C \ \ \ \). I'll give a translation to your headline. "Do the antiderivatives of a function, where the function is equal to 1 divided by an expression, equal the natural logarithm of that expression (plus some arbitrary constant)?" Look at \(\displaystyle \ \ \int \frac{1}{2} dx = \frac{1}{2}x \ + \ C. \ \ \ \ \ \) Or look at \(\displaystyle \int \frac{1}{x^2}dx \ = \ \int x^{-2}dx \ = \ -\frac{1}{x} \ + \ C\) \(\displaystyle \ \ \ \ \ \ Also, \ \ this \ \ question \ \ belongs \ \ in \ \ a \ \ calculus\ \ subforum.\)

 

[Deleted member 4993]

example: (if you wanted to solve for 'x')

(1) \(\displaystyle \dfrac{1}{x} = \ln (x)\)

(2) \(\displaystyle \dfrac{1}{4x - 3} = \ln (4x - 3)\)

For (1), x = ~ 1.76322 ... use Newton's method

 

[Jason76]

Ok I see, the integral of 1 over a math expression = the natural log of the math expression. To go further, if you make ln math expression a power of e, then you get the math expression by itself (without ln).

 

[MarkFL]

You should be aware that in general:

\(\displaystyle \displaystyle \int\frac{1}{f(x)}\,dx\ne\ln|f(x)|+C\)

while:

\(\displaystyle \displaystyle \int\frac{\frac{df}{dx}}{f(x)}\,dx=\ln|f(x)|+C\)