Question

In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b,$ for some fixed $b \ge 2.$ A Martian student writes down $3\log(\sqrt{x}\log x) = 56$ $\log_{\log x}(x) = 54$ and finds that this system of equations has a single real number solution $x \gt 1.$ Find $b.$

Answer

Answer: 216

Concepts:logarithm system of equations substitution

Difficulty rating: 2400

Solution:

Let $y = \log_b x.$ By the change-of-base formula, $\begin{aligned} \log_{\log x}(x) &= \frac{\log_b x}{\log_b(\log_b x)} \\ &= \frac{y}{\log_b y} = 54, \end{aligned}$ so $\log_b y = \frac{y}{54}.$ The first equation says $3\left(\frac{1}{2}\log_b x + \log_b(\log_b x)\right) = 56,$ that is, $\frac{y}{2} + \log_b y = \frac{56}{3}.$

Substituting $\log_b y = \frac{y}{54}$ gives $\frac{y}{2} + \frac{y}{54} = \frac{56}{3},$ so $\frac{28y}{54} = \frac{56}{3}$ and $y = 36.$ Then $\log_b 36 = \frac{36}{54} = \frac{2}{3},$ so $b^{2/3} = 36$ and $b = 36^{3/2} = 216.$

2019 AIME II Problem 6

Answer: 216

Difficulty rating: 2400

Solution:

Problem 6 in Other Years