Why $\log xy=\log x+\log y$? It is of course well known and basic formula. I am just curious. Is there a proof for it?
How to prove that $\log xy=\log x+\log y$?
 A: We have $$\ln (xy) = \int_{1}^{xy}\frac{dt}{t}=\int_{1}^{x}\frac{dt}{t}+\int_{x}^{xy}\frac{dt}{t}=\ln(x)+\int_{x}^{xy}\frac{dt}{t}.$$ For the last integral, we substitute $u=\frac{t}{x}$ to get $du=\frac{dt}{x}$, thus reducing the last integral to $\int_{1}^{y}\frac{du}{u}=\ln (y)$. Hence, $\ln(xy)=\ln(x)+\ln(y)$.
Dividing both sides by $\log (a)$, we have $\log_{a}(xy)=\log_{a}(x)+\log_{a}(y)$ for any base $a\neq 1.$
A: $$e^{\ln(x)+\ln(y)}=e^{\ln(x)}e^{\ln(y)}=xy$$ so $$\ln(xy)=\ln(x)+\ln(y)$$
As noted in the comments, this works for any log. I used base $e$ out of habit/convenience.
A: Let $m = \log_a(x)$ and let $n = \log_a(y)$
Express $x$ and $y$ in terms of exponents, so $x = a^m$ and $y = a^n$.
Therefore $xy = a^{m+n}$
Take the $\log$ of both sides to obtain:
$$\log_a(xy) = \log_a(a^{m+n})$$
$$\log_a(xy) = (m+n)*\log_a(a)$$
$$\log_a(xy) = m + n$$
$$\log_a(xy) = \log_a(x) + \log_a(y)$$
A: Under the definition of $\ln(x)$ by integration of $\frac{1}{x}$, there is another way to work with the integral 
$$\int_{x}^{xy}\frac{dt}{t}$$
By splitting the integral into
$$\int_{x}^{xy}\frac{dt}{t}=\int_{1}^{xy}\frac{dt}{t}-\int_{1}^{x}\frac{dt}{t}.$$
This follows readily from the fact that the integral is in essence a signed area under a curve, or more formally, from the Riemann sum.
The rest is the same as the proof given by user Indrayudh Roy.
A: The logarithm of a number with respect to a certain base is just the exponent to which this vase must be raised to give the number. Thus, if we express $N>0$ as a power of a base $1\ne b>0,$ we have that $$b^{\ell}=N.$$ This $\ell$ is what is called the logarithm of $N$ to the base $b,$ usually written $\log_bN$ for short.
Thus, suppose we have two numbers expressed as powers of the same base $b,$ say $b^x$ and $b^y,$ then $x,y$ are respectively the logs of these numbers to their common base, $b.$ Now if you multiply these numbers you get (according to the basic rule of indices) $$b^xb^y=b^{x+y},$$ so that the logarithm of the product to the base $b$ is just the sum of the logarithms $x+y.$
