Weekly Challenge 43: A Close Match

Can you massage the parameters of these curves to make them match as closely as possible?

Weekly Challenge 44: Prime Counter

A weekly challenge concerning prime numbers.

Weekly Challenge 28: the Right Volume

Can you rotate a curve to make a volume of 1?

Weekly Challenge 24: Suspicious Integrator

Stage: 5 Challenge Level:

The integrator (with proper formatting) gave the answer for my integral $I$ as

$$I = \frac{25000\log[-32(99999 + 16^x)]}{\log 2}$$

The problem is that the logarithm of a negative number is complex. However, the rules of logarithms still apply, so we see that
$$I = \frac{25000}{\log 2}\Big(\log(-32) + \log(99999+16^x)\Big)$$
The $\log(-32)$ part, although complex, is a constant of integration which has been chosen by the integrator. Thus, the integral is of the form
$$I = \frac{25000\log(99999+16^x)}{\log 2} + c$$
This is perfectly real for real choices of $c$ and direct computation shows that this differentiates down to our starting function.