# Logarithmic lawsapplication

1. Jan 12, 2014

### 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.....!

2. Jan 12, 2014

### Mentallic

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

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

3. Jan 12, 2014

### 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!

4. Jan 12, 2014

### Mentallic

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

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?