Given two values $a$ and $b$, how should one go about solving the following inequality for $r$:
$$a \le b^r -r .$$
Applying $\log_b$ on both sides of the inequality doesn't help me much since that yields the following:
$$\log_b a \le \log_b (b^r-r) .$$
I know that
$$\log_z x - \log_z y = \frac{\log_z x}{\log_z y} .$$
But that doesn't help me to eliminate the exponentiation and solve for $r$.
It's been ages that I've done algebra, and now I'm back in grad school, and I'm finding this equation in a homework. Can't remember ever seeing a rule of logarithms involving such expressions.
The actual homework (after simplification) involves $9 \le 2^r - r$ which is trivially solvable by just eyeballing it.
But is there a methodical way to solve $a \le b^r-r$ for $r$ given any arbitrary numbers $a$ and $b$?