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It is of course well known and basic formula. I am just curious. Is there a proof for it? How to prove that $\log xy=\log x+\log y$?

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

5 Answers 5

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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 $a\neq 1.$

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  • $\begingroup$ 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... $\endgroup$
    – user140943
    Apr 7, 2014 at 21:02
  • $\begingroup$ @user140943 , the OP has himself tagged calculus. $\endgroup$ Apr 8, 2014 at 6:40
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$$e^{\ln(x)+\ln(y)}=e^{\ln(x)}e^{\ln(y)}=xy$$ 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.

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  • $\begingroup$ 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 $\endgroup$
    – Mathias711
    Apr 7, 2014 at 20:30
  • $\begingroup$ @Mathias711: $e$ in these formulas is not an "exponent" but a "base". That's why it's the base of the logarithm! $\endgroup$ Apr 7, 2014 at 20:35
  • $\begingroup$ @HenningMakholm Yeah, that is what I meant. I couldn't come up with the word. Thanks for clarifying. $\endgroup$
    – Mathias711
    Apr 7, 2014 at 20:58
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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: $$\log_a(xy) = \log_a(a^{m+n})$$ $$\log_a(xy) = (m+n)*\log_a(a)$$ $$\log_a(xy) = m + n$$ $$\log_a(xy) = \log_a(x) + \log_a(y)$$

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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.

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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 $x+y.$

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