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 $b$?


The case of equality can be solved in terms of the Lambert W function.

It can be used to solve equations of the form $$p^{ax+b} = cx + d$$

(quoted from the wiki page).

$b^{r} -r$ is 'mostly' monotonic, so I suppose solving the equality will be enough to find the solutions to your inequality.

  • $\begingroup$ Holy cow, I'll have to brush on that one. Thanks! $\endgroup$ – luis.espinal Oct 19 '10 at 18:06

You can also get a quick-and-dirty approximation for the equality case with the following approach, which should hold in most situations when $a, b > 1$.

Rearrange the problem and take logarithms to rewrite it as $$\ln\left(1+ \frac{r}{a}\right) = r \ln b - \ln a.$$

Now, if $\frac{r}{a}$ is less than 1 (and it should be in most cases where $a, b > 1$),

$$\ln\left(1+ \frac{r}{a}\right) \approx \frac{r}{a},$$

which gives you $$r \approx \frac{- \ln a}{1/a - \ln b}.$$

For your specific problem, this method yields $r \approx 3.78$, whereas the actual answer is closer to $3.66$.

You can improve this by using the quadratic approximation

$$\ln\left(1+ \frac{r}{a}\right) \approx \frac{r}{a} - \frac{r^2}{2a^2},$$ which requires solving a quadratic equation.

(These approximations come from the Taylor expansion of $\ln (1+x)$.)


If you don't have the W function, you need an iterative root finding type of solution. Note that for your example r is somewhere between 3 and 4. This equation is nicely behaved, so any of them will work. See Root finding for a start

  • $\begingroup$ Hmmm, I was suspecting I'll have to use an iterative solution, but I was hoping to find a solution involving some form of algebraic manipulation. Oh well, at least now I'm on the right track. Thanks! $\endgroup$ – luis.espinal Oct 19 '10 at 18:07

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