Thanks for that Rui!!
Another one that makes no sense... I got an answers but it was wrong and I don't quite understand what they did (part ii)
Thanks again
Sure! So let's start by doing what we always do to find tangents to curves, differentiate. I'll call the two graphs \(y_1\) and \(y_2\):
So we'll come back to that later; we also need the point of intersection, and at that point of intersection:
But remember the gradients from above will need to be equal too (common tangent), so we also have \(re^{rx}=\frac{1}{x}\).
What we have here are two equations linking \(r\) and \(x\); we're going to solve these simultaneously! The goal will be to use the result we were just given somewhere too. With that in mind, let's rearrange one of those equations:
Aha, theres the result from Part (i); we can then say that \(rx\approx0.56\)!
Therefore back in \(*\), we have: \(\log_e{x}=e^{0.56}\), and that's an equation you can solve for \(x\), and then substitute back to find \(r\); does that make sense?