Showing that ln$(xy)$ = ln $x$+ln $y$
Let ln $x\ =\ \int_1^x \frac{1}{t}\ dt.$
How do I show that ln$(xy)$ = ln $x$+ln $y$ where $x$ and $y$ are positive reals. I read the following proof from Limaye book.
Fix $y\in (0,\infty).$ Consider $f(x)=$ ln $xy$ -ln $x$. Then $f'(x)= \frac{1}{xy}.y-\frac{1}{x}=0 \ \forall\ x \in (0,\infty). $
$\therefore f$ is constant. $f(x)=f(1)=$ ln $y$ - ln $1 = $ ln $y.$
$ \therefore f(x) = $ln$ (xy) - $ln$ x = $ln$ y\ \implies\ $ln$ (xy) = $ln$ \ x + $ ln$\ y$
I feels that this proof is so constructive.
Im wondering to know some alternative proofs.
Thanks in Advance...
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$\begingroup$We have $$f(xy) = \int_1^{xy}\dfrac{dt}t = \underbrace{\int_1^x \dfrac{dt}t+ \int_x^{xy} \dfrac{dt}t = \int_1^x \dfrac{dt}t + \int_1^y \dfrac{dz}z}_{\text{Setting $z=t/x$}} = f(x) + f(y)$$
$\endgroup$ 1 $\begingroup$The correct solution depends on your definition of $\ln(x)$.
One definition is: $\ln(x)$ is the unique real number satisfying $e^{\ln(x)} = x$.
Now: $e^{\ln(x) + \ln(y)} = e^{\ln(x)}e^{\ln(y)} = xy = e^{\ln(xy)}$. So we must have $\ln(x) + \ln(y) = \ln(xy)$.
$\endgroup$ 2 $\begingroup$Let $x,y > 0$. Define the following functions: $$f_y(x)=\ln(xy) = \int_1^{xy} \frac{1}{t} dt$$
$$g_y(x)=\ln(x)+\ln(y) = \int_1^x \frac{1}{t} dt + \int_1^y \frac{1}{t} dt$$
Now we take their derivatives to find: $$\frac{d}{dx} f_y(x) = y \frac{1}{xy}=\frac{1}{x}$$ by the fundamental theorem of calculus. Also $$\frac{d}{dx} g_y(x) = \frac{1}{x}.$$
Note that $f_y(1) = \ln(y) = g_y(1)$. Thus $f_y(x)=g_y(x)$ for all $x$, since the two functions have the same derivative and agree at one point.
$\endgroup$ $\begingroup$As you asked for a hint ( good job) If you see the definition of $\ln(x)$ you can notice that it's the inverse function of $ e^x$ then follow the law of exponents and take the inverse function on both sides. For example ,to prove$ cos^{-1}x=sin^{-1}(1-x^2)^{1/2} $ We have $$ sin(\theta)=(1-(cos(\theta)^2))^{1/2}$$ Put, $$cos(\theta)=x$$ we get $$ cos^{-1}x=sin^{-1}((1-x^2)^{1/2}) $$
$\endgroup$ $\begingroup$Integrals are ugly. I prefer this elegant solution method:
$\ln(xy)=\ln(x)+\ln(y)$ iff $\exp(\ln(xy))=\exp(\ln x+\ln y)$ iff $xy=\exp(\ln x)\exp(\ln y)$ iff $xy=xy$.
$\endgroup$ 2 $\begingroup$Using the Definition ln(x)$\ =\ \int_1^x \frac{1}{t}\ dt$, Construct the function $$f(x)= ln(xy)-ln(x)$$ where $ x,y >0$
Therefore, $$f'(x)= \frac{1}{x}-\frac{1}{x}=O$$
Meaning that $f(x)$ is a constant function, so $ln(xy)-ln(x)=c$ where $c$ is some constant. Let $x=1$ Then $$f(x)=ln(y)-ln(1)=c \implies c=ln(y)$$
Therefore $$ln(xy)-ln(x)=ln(y) \implies ln(xy)=ln(x)+ln(y)$$
Which is what you wanted to prove
$\endgroup$ $\begingroup$Let $p_1,p_2>0$.
$p_1p_2=e^{r_1}e^{r_2}=e^{r_1+r_2}$, $r_1,r_2\in\mathbb{R}$.
So, $\ln(p_1p_2)=r_1+r_2=\ln(p_1)+\ln(p_2)$.
I ignored uniqueness here. Constructive criticism is welcomed.
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