Define the natural logarithm of a number x in terms of an integral.Show that
ln(xy) = ln(x) + ln(y)
for all x,y >0
define $\displaystyle \ln(x)=\int_{1}^{x}\frac{1}{t}dt$
we want to show $\displaystyle \ln(xy)=\ln(x)+\ln(y)$
so we start with
$\displaystyle \ln(xy)=\int_{1}^{xy}\frac{1}{t}dt$
using the linear property of integrals we get
$\displaystyle \int_{1}^{xy}\frac{1}{t}dt=\int_{1}^{x}\frac{1}{t} dt+\int_{x}^{xy}\frac{1}{t}dt$
lets focus on the 2nd integral
$\displaystyle \int_{x}^{xy}\frac{1}{t}dt$
let $\displaystyle u=\frac{t}{x} \iff ux =t$
$\displaystyle du=\frac{1}{x}dt \iff xdu=dt$
$\displaystyle \int_{x}^{xy}\frac{1}{t}dt=\int_{1}^{y}\frac{1}{ux }xdu=\int_{1}^{y}\frac{1}{u}du=\ln(y)$
so finally with all the above work
$\displaystyle \ln(xy)=\int_{1}^{xy}\frac{1}{t}dt=\int_{1}^{x}\fr ac{1}{t}dt+\int_{1}^{y}\frac{1}{u}du=\ln(x)+\ln(y)$
Yeah!!
No it is not. How is $\displaystyle e^x$ defined? In fact, $\displaystyle e^x$ is defined to be the inverse function of $\displaystyle \ln x$ but you never demonstrated it.
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Here is how to prove it. Let $\displaystyle y>0$. Define a function $\displaystyle f0,\infty)\mapsto \mathbb{R}$ as $\displaystyle f(x) = \ln (xy) - \ln x$. This function is differenciable and furthermore, $\displaystyle f'(x) = \frac{y}{yx} - \frac{1}{x} = 0$. Thus, $\displaystyle f(x) = k$ for some number $\displaystyle k$. In particular, $\displaystyle f(1) = k$ but $\displaystyle f(1) = \ln (y) - \ln (1) = \ln y$. Thus, $\displaystyle k=\ln y$. Therefore, $\displaystyle \ln (xy) - \ln x = \ln y$.