Why is the integral characterization of e the same as the derivative characterization of e?

Question

Why is it the case that $$\frac{d}{dx}b^{x}=b^x \iff b = e$$ is equivalent to $$\int_1^b{\frac{1}{t}dt}=1 \iff b = e?$$

Answer 1

If you restrict $x > 0$, the differential equation is also true if $b = 0$. Putting that case aside and since $b^x$ is not defined for $b< 0$ and non-integer $x$, assume $b > 0$.

Let $y(x) = b^x$ and note that $y(0) = 1, y(1) = b$. So the differential equation becomes $\frac{dy}{dx} = y$, or $$\frac 1y\frac{dy}{dx} = 1$$ Also note that since $y$ is differentiable, it is continuous, and by the differential equation, $\frac{dy}{dx}$ is also continuous.

Integrate both sides: $$\int_0^1\frac1y\frac{dy}{dx}\,dx = \int_0^1 1\,dx = 1$$ Make the substitution $t=y, dt = \frac{dy}{dx}dx, x= 0 \implies t = 1, x=1 \implies t = b$: $$\int_1^b\frac{dt}t = 1$$

The inverse implication follows from defining $f(y) = \int_1^y\frac{dt}t$ and differentiating.