Why $\log(x+1)=\log(1+\frac{1}{x})+\log(x)$?

Question

Why is this expression true? $\log(x+1)=\log(1+\frac{1}{x})+\log(x)$

What logarithmic property allows the equivalence?

Answer 1

Recall that

$$\log a + \log b = \log ab$$

therefore

$$\log\left(1+\frac{1}{x}\right)+\log(x)=\log(x+1)$$

To prove that foundamental property recall that by definition

• $A=\log a \iff e^A=a $

• $B=\log b \iff e^B=b $

then

$$ab=e^A\cdot e^B = e^{A+B} \iff \log ab = \log a + \log b$$

which clearly shows that the property is related to the corresponding property for exponential function.

Answer 2

If you just start with the expression to the right-hand side you can get the left-hand side as follows:

$\log (x+1)=\log((x+1)\displaystyle\frac{x}{x})=\log[(1+\frac{1}{x})x]=\log(1+\frac{1}{x})+\log x$

The similar technique will work to get the left-hand side from the right-hand side expression as well.