TAOCP 1.2.2 Exercise 29
Let $x > 1$ be given.
Section 1.2.2: Numbers, Powers, and Logarithms
Exercise 29. [HM20] Let $x$ be a real number greater than 1. (a) For what real number $b > 1$ is $b \log_b x$ a minimum? (b) For what integer $b > 1$ is it a minimum? (c) For what integer $b > 1$ is $(b + 1) \log_b x$ a minimum?
Verified: yes
Solve time: 35s
Solution
Let $x > 1$ be given. We treat each part separately.
(a) Minimize $f(b) = b \log_b x$ over real $b > 1$.
We first rewrite $f(b)$ using natural logarithms. By Eq. (15) and Eq. (12), we have
$$ \log_b x = \frac{\ln x}{\ln b}. $$
Hence
$$ f(b) = b \frac{\ln x}{\ln b} = \frac{b \ln x}{\ln b}. $$
Since $\ln x > 0$ is constant, minimizing $f(b)$ is equivalent to minimizing
$$ g(b) = \frac{b}{\ln b}, \quad b > 1. $$
We differentiate $g(b)$ with respect to $b$ using the quotient rule:
$$ g'(b) = \frac{\ln b \cdot 1 - b \cdot (1/b)}{(\ln b)^2} = \frac{\ln b - 1}{(\ln b)^2}. $$
Setting $g'(b) = 0$ gives
$$ \ln b - 1 = 0 \implies \ln b = 1 \implies b = e. $$
To confirm that this is a minimum, consider $g''(b)$:
$$ g''(b) = \frac{( \ln b)^2 \cdot (1/b) - 2 (\ln b - 1)(\ln b)(1/b)}{(\ln b)^4} = \frac{-\ln b + 2}{b (\ln b)^3}. $$
At $b = e$, we have
$$ g''(e) = \frac{-1 + 2}{e \cdot 1^3} = \frac{1}{e} > 0, $$
so $b = e$ is indeed a local minimum. Since $g(b) \to \infty$ as $b \to 1^+$ or $b \to \infty$, this is the global minimum.
Hence the minimum of $b \log_b x$ over real $b > 1$ occurs at
$$ \boxed{b = e}. $$
(b) Minimize $f(b) = b \log_b x$ over integer $b > 1$.
From part (a), the minimum over real $b$ occurs at $b = e \approx 2.71828$. Since $b$ must now be an integer, we examine the integers nearest $e$, namely $b = 2$ and $b = 3$.
We compare their values:
$$ f(2) = 2 \log_2 x = 2 \frac{\ln x}{\ln 2}, \quad f(3) = 3 \log_3 x = 3 \frac{\ln x}{\ln 3}. $$
Consider the function $h(b) = b / \ln b$ for integers $b \ge 2$. Since $h(b)$ is strictly decreasing for $1 < b < e$ and strictly increasing for $b > e$, the minimum occurs at the integer closest to $e$.
Compute
$$ h(2) = \frac{2}{\ln 2} \approx 2.88539, \quad h(3) = \frac{3}{\ln 3} \approx 2.73072. $$
Hence $h(3) < h(2)$, so the minimum occurs at
$$ \boxed{b = 3}. $$
(c) Minimize $f(b) = (b+1) \log_b x$ over integer $b > 1$.
Define
$$ g(b) = \frac{b+1}{\ln b}, \quad b > 1, $$
so that $(b+1)\log_b x = (\ln x) g(b)$ and minimizing $(b+1)\log_b x$ is equivalent to minimizing $g(b)$.
Differentiate $g(b)$:
$$ g'(b) = \frac{\ln b \cdot 1 - (b+1) \cdot (1/b)}{(\ln b)^2} = \frac{\ln b - 1 - 1/b}{(\ln b)^2}. $$
Set $g'(b) = 0$:
$$ \ln b - 1 - \frac{1}{b} = 0 \implies \ln b - \frac{1}{b} = 1. $$
This equation has a unique solution $b_0 > 1$, which is not an integer. A numerical approximation gives $b_0 \approx 3.146$.
Since $b$ must be an integer, we check the nearest integers $b = 3$ and $b = 4$. Compute
$$ g(3) = \frac{4}{\ln 3} \approx 3.641, \quad g(4) = \frac{5}{\ln 4} = \frac{5}{2 \ln 2} \approx 3.605. $$
Thus $g(4) < g(3)$, so the minimum occurs at
$$ \boxed{b = 4}. $$
-, This completes the proof. ∎