# 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 $\log xy=\log x+\log y$?

• It depends on how you define the logarithm. Apr 7, 2014 at 20:27
• Some people may claim that the motivation of defining the logarithm is the functional equation $f(xy)=f(x)+f(y)$. Apr 7, 2014 at 20:32
• Still have my old slide rule... Apr 7, 2014 at 21:09
• What side rule? Apr 7, 2014 at 21:56

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 $$a\neq 1.$$

• I like this because it's cool, but I don't think this would work for the average algebra student unless they happen to know calculus... Apr 7, 2014 at 21:02
• @user140943 , the OP has himself tagged calculus. Apr 8, 2014 at 6:40

$$e^{\ln(x)+\ln(y)}=e^{\ln(x)}e^{\ln(y)}=xy$$ 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.

• And this of course applies to every log, but you have to change the exponent ($e$ in the example) by the base of the logarithm. Then its proven for all cases Apr 7, 2014 at 20:30
• @Mathias711: $e$ in these formulas is not an "exponent" but a "base". That's why it's the base of the logarithm! Apr 7, 2014 at 20:35
• @HenningMakholm Yeah, that is what I meant. I couldn't come up with the word. Thanks for clarifying. Apr 7, 2014 at 20:58

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: $$\log_a(xy) = \log_a(a^{m+n})$$ $$\log_a(xy) = (m+n)*\log_a(a)$$ $$\log_a(xy) = m + n$$ $$\log_a(xy) = \log_a(x) + \log_a(y)$$

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.

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 $$x+y.$$