There is a number which it's natural logarithm is equal to 1

Prove that there is a number $e$ at which the area under the curve of $\frac{1}{x}$ is equal to 1, and  show that the number $e$ should be $2<e<3$;

It is obvious from the graph picture that we can have an area of the size of 1;
first of all we check x=2,we can draw a rectangle from x=1 to x=2, because the equation is 1/x, at x=1 the equation gives us y=1=1/1 which is the high of our rectangle(in this case it would be square). the area of our shape(gray region) would be 1*1=1; and we can see that the actual area of the curve at this region is less than our square(because of concavity); therefore at x=2 we can not have the area 1 and x must be more than 1;
now we have to prove that the x can not be more than 3. if we draw tangent line of equation 1/x at x=2, and then calculate the area under our tangent line equation we can judge about e.
the line equation at x=2 would be :
$$
f(x)=\frac { 1 }{ x } ,\quad f(x)'=-\frac { 1 }{ x^{ 2 } } ,\quad f(2)'=-\frac { 1 }{ 4 } \\ \\ l(x)=f(2)+f(2)'(x-2)\\ l(x)=\frac { 1 }{ 2 } -(\frac { 1 }{ 4 } )(x-2)
$$
the area of the l(x) from 1 to 3 gives us the trapezoid, we can calculate the area in two way, using classic geometry or definite integral, we do it by the second method:

$$
\\ \int _{ 1 }^{ 3 }{ l(x)dx } =\int _{ 1 }^{ 3 }{ [\frac { 1 }{ 2 } -(\frac { 1 }{ 4 } )(x-2)]dx } =[x-\frac { 1 }{ 8 } x^{ 2 }]\overset { 3 }{ \underset { 1 }{ \quad  }  } =1\\
$$
the area of the trapezoid is equal to 1, but we can see from the picture that the area of actual curve in this interval is more than the trapezoid area. the exact area is $ trapezoid area(=1) + \Delta s$. at the points greater than 3 the area must be more than 1; therefore $e$ should be less than 3;

Notice that the number e is very useful, we know some important properties of natural logarithm like: $ln(a^{x})=x.ln(a)$ (we can prove it easily by the fact that the functions  $ln(ax)$ and $ln(x)$ have the same derivative), now we have a substitute for b at which the ln is equal to 1, namely $ln(e)=1$. therefore we achieved something wonderful! the inverse function of $ln(x)$. remember that two inverse function have this property : $f(f(x))^{-1}=x$. now we have this property!
$ln(e^{x}) =x.ln(e)=x$ , therefore the functions $e^{x}$ and $ln(x)$ are inverses of each other.