# proof of $\log(xy) =\log (x) + \log (y)$ [closed]

$\log(xy) = \log (x) + \log ( y)$ and its division counter part were mentioned in an axiomatic way which I failed to proof.

I noted the correlation with the exponential rules of adding powers in case of multiplication of a single base and subtracting them in case of division but if we do proceed with that, how do we take log of terms linked with arithmetic operators ?

it doesn't have to be base $10$ but I picked it for convenience.

another question , using the fact that y = log x to the base a is the inverse function of x = a^y does that mean that the identitiy mentioned at the begingin is the inverse of 10^x+y = 10^x times 10 ^ y ?

• Yes, do 10^ on both sides – Simply Beautiful Art Nov 6 '16 at 23:44
• Before you can prove anything about $\log$ you need a definition of $\log$. – GEdgar Nov 6 '16 at 23:54
• What's up with close votes? The question is (1) clear (2) has context. There may be a duplicate somewhere but that's not why people seem to be voting. – MathematicsStudent1122 Nov 7 '16 at 7:12
• This is direct result of the definition of $\log_b z$ being defined as the number, $y$ so that $b^y = z$. $\log_b xy$ is the $a$ so that $b^a = xy$ $\log_b x$ is the $c$ so that $b^c = x$ and $\log_b y$ is the $d$ so that $b^d = y$ so $b^a = xy = b^cb^d = b^{c+d}$. The number require to raise b so that it equals xy, is the sum of the numbers required to raise b to get x and to get y. – fleablood Nov 7 '16 at 21:47
• "another question , using the fact that y = log x to the base a is the inverse function of x = a^y does that mean that the identitiy mentioned at the begingin is the inverse of 10^x+y = 10^x times 10 ^ y ?" Syntactically, I don't think that sentence makes any sense. But if you mean, what I think you mean the yes. $10^{\log x} = x$ so $10^{\log x + \log y} = 10^{\log x}*10^{\log y} = x*y = 10^{\log xy}$. – fleablood Nov 7 '16 at 21:52

I want to take a geometric view here. I promise it won't be too bad! Let's start with the fact that, with a base $b$,

$b^{n}b^{m} = b^{n+m}$

Which makes sense just from thinking about groupings of numbers,

$(b\cdot b \cdots b)_{ntot}\cdot(b \cdot b \cdots b)_{mtot} = (b\cdot b \cdots b)_{ntot+mtot}$.

Let's define a function $f(x)$ to be,

$f(x)=b^x$.

Then we have the properties,

$f(n+m) = b^{n+m} = b^n b^m = f(n)f(m)$,

from what I discussed in the first sentence. But what does this actually mean? If we look at a graph (which I've crudely drawn because I am not good with drawing software),

This shows graphically what is happening with $n$ and $m$ and how the function responds (sorry for high quality?). But what about $log_b (x)$? Let's define a new function

$g(x) = log_b (x)$.

It turns out that $log_b(x)$ and $b^x$ are related by a reflection about the line $y = x$ on a Cartesian grid. This is shown very nicely in this picture:

https://people.richland.edu/james/lecture/m116/logs/log2.gif

If you imagine the line $y = x$ going straight through the origin you'll see that they are just reflections of each other! What does this have to do with your question? Well what if we imagine taking the $x$ axis in my original plot, and treating that as our new '$y$' axis (vertical axis), with the $f(x)$ axis being the $g(x)$ axis (horizontal axis). I've drawn another sketch (rotated by 90 degrees counterclockwise on right):

Now usually we like our positive direction to be to the right, so we'll flip it and I'll replot $n$ and $m$ on this graph:

Since it was just a reflection, we're seeing another exponentially increasing function! $n$ and $m$ are on our vertical 'x' axis and $g(n)$ and $g(m)$ are on our horizontal '$g(x)$' axis. But we know on the vertical axis the two values multiply to give a new value, then on the horizontal axis those two numbers add to give the final value for exponentially increasing functions! All we've done is swapped the axis and used the properties of $f(x)=b^x$ to show that,

$g(n)+g(m) = g(nm)$

For $g(x)=\log_b(x)$! Plugging in,

$\log_b(n) + \log_b(m) = \log_b(nm).$

• appreciate this answer even tho it took time to be intuitive to +1 for using plain functions – sarah Nov 7 '16 at 21:30

Let $r=\log(x), s=\log(y)$ and $t=\log(xy)$.

Then $x=10^r,y=10^s$ and $xy=10^t=10^r10^s=10^{r+s}$.

Therefore, $t=r+s$, that is, $\log(xy)=\log(x)+\log(y)$.

• +1 for no calculus. Plain and simple is all it needs to be. – Fine Man Nov 7 '16 at 6:22

Using a different definition of the logarithm, $\ln(x) := \int_1^x \frac{dt}{t}$, we can prove this without a definition of exponentiation.

$$\ln(ab) = \int_1^{ab}\frac{dt}{t} = \int_1^a\frac{dt}{t} + \int_a^{ab} \frac{dt}{t} = \ln(a) + \int_a^{ab} \frac{dt}{t}$$ Next we can do a u-substitution with $u= \frac ta$, meaning the limits of the remaining integral can be re-expressed as $1$ and $b$. Also, note that $a du = dt$ as well as $au = t$, meaning we find that $\int_a^{ab} \frac{dt}{t} = \int_1^b \frac{a\,du}{au} = \int_1^b \frac{du}{u} = \ln(b)$.

Now we find that $\ln(ab) = \ln(a) + \ln(b)$, using just basic properties of integrals.

Considering that $\log(x)$ is the inverse function of $e^x$ you have

$$e^{\log(xy)} = xy = e^{\log(x)} \cdot e^{\log(y)} = e^{\log(x)+\log(y)}$$

But since the exponential function is injective you get $\log(xy) = \log(x)+\log(y)$

• Personally, I'd just go with log base $b$ for those who haven't seen $e$. – Simply Beautiful Art Nov 7 '16 at 0:04
• @SimpleArt I suppose if someone hasn't seen $e$, it could just as well be an arbitrary base. – David Z Nov 7 '16 at 4:40
• @SimpleArt Isn't it unlikely that someone has seen enough of $log$ to know that $log(xy)=log(x)+log(y)$, but has not seen $e$ yet? – Right Leg Nov 7 '16 at 5:02
• i m sorry , i used log in the maths way where it stands for logarithm of base ten not as the natural logarithm , as to the answer , i will check if i understand it . – sarah Nov 7 '16 at 21:24
• @sarah well, then $\log(x)$ is just the inverse of $10^x$ – cronos2 Nov 7 '16 at 21:25

$\log xy = \log x + \log y\\ 10^{\log xy} = 10^{(\log x + \log y)}\\ 10^{\log xy} = (10^{\log x})(10^{\log y})\\ xy = (x)(y)$

Let $b, x, y > 0$, with $b \neq 1$. Suppose that $log_b x = u$ and that $\log_b y = v$. Then, by definition of the logarithm, $b^u = x$ and $b^v = y$. Therefore, $$\log_b{xy} = \log_b(b^ub^v) = \log_b b^{u + v} = u + v = \log_b x + \log_b y$$ and $$\log_b\left(\frac{x}{y}\right) = \log_b \left(\frac{b^u}{b^v}\right) = \log_b b^{u - v} = u - v = \log_b x - \log_b y$$