Solving equations involving exponential and line

Question

I understand how to do (a): y=(p-1)x/(lnp) +-p-1 but I have no idea where to start in (b) and how to solve it. Where did the k come from?

Answer 1

The correct answer for part (a) is $$y(x) = \frac{p-1}{\ln(p)}x + p - 1.$$

This is easily found by finding the $x$ and $y$ intercepts and using the equation for a line between two points.

For part (b), we can rearrange the equation to obtain $$\frac{e^k - 1}{k}x = -e^{-x} + 1.$$ Let $g(x)$ be the LHS and $h(x)$ be the RHS. Note that the slope of $g$ is equal to the slope of $y$ with $p = e^k$, and $h$ is just $f$ with $p = 1$. Also note that $h(0) = 0$.

Side note: This observation makes it clear why part (a) had the condition $p > 1$. If $p = 1$, the $x$ and $y$ intercepts are the same point, so there is no unique line that intersects both.

Let $p = e^k$ and observe that $g(x) = y(x) - p + 1$ and $h(x) = f(x) - p + 1$. Remember that we are trying to solve $g(x) = h(x)$. Using this result, we transform that problem into solving $y(x) = f(x)$.

By the definition of $y$, If $y(x) = f(x)$, then $x = 0$ (the $y$ intercept) or $y(x) = 0$ (the $x$ intercept). Solving $y(x) = 0$, we obtain $x = -k$. Therefore, the solutions are $x = 0$ and $x = -k$.

Side note: Regrettably, this answer does not really show the process by which I discovered the approach, which primarily involved graphing things to visualize the problem and looking for patterns. Either way, I hope it helps.