Why is my computer giving different answers for logarithmic laws?

[alingy1]

I'm using this computer program:
It tells me that:
log(x+1/3)=log(3x+1)
Why is that so? I plug in values: they give different answers...!

 

[Mentallic]

I'm using this computer program:
It tells me that:
log(x+1/3)=log(3x+1)
Why is that so? I plug in values: they give different answers...!

\log(3x+1)=\log(3(x+1/3))=\log(3)+\log(x+1/3)

Since \log(3)\neq 0 for any base of log, something more elusive must be happening.

 

[alingy1]

The reason I am asking this is that I am trying to calculate the integral of 1/(3x+1).
I found it equals (1/3)(integral of (1/(x+1/3))=1/3ln((x+1/3)). But, the computer says it is wrong!

 

[Mentallic]

The reason I am asking this is that I am trying to calculate the integral of 1/(3x+1).
I found it equals (1/3)(integral of (1/(x+1/3))=1/3ln((x+1/3)). But, the computer says it is wrong!

\frac{d(\log{x})}{dx}=\frac{1}{x}

only applies when the base of log is e (usually denoted ln(x)). In general,

\frac{d(\log_{a}{x})}{dx}=\frac{1}{x\ln{a}} since \log_{a}{x}=\frac{\ln{x}}{\ln{a}} hence \frac{d(\log_{a}{x})}{dx} = \frac{d(\frac{\ln{x}}{\ln{a}})}{dx}=\frac{1}{\ln{a}}\frac{d(\ln{x})}{dx}=\frac{1}{\ln{a}}\cdot\frac{1}{x}

Maybe your computer is using a base other than e?