2019 AIME II Problems/Problem 6

Problem 6

In a Martian civilization, all logarithms whose bases are not specified as assumed to be base $b$ , for some fixed $b\ge2$ . 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 $x>1$ . Find $b$ .

Solution

Using change of base on the second equation to base b, $\frac{\log x}{\log \log x }=54$ $\log x = 54 \cdot \log \log x$ $b^{\log x} = b^{54 \log \log x}$ $x = (b^{\log \log x})^{54}$ $x = (\log x)^{54}$ Substituting this into the $\sqrt x$ of the first equation, $3\log((\log x)^{27}\log x) = 56$ $3\log(\log x)^{28} = 56$ $\log(\log x)^{84} = 56$

We can manipulate this equation to be able to substitute $x = (\log x)^{54}$ a couple more times: $\log(\log x)^{54} = 56 \cdot \frac{54}{84}$ $\log x = 36$ $(\log x)^{54} = 36^{54}$ $x = 6^{108}$

However, since we found that $\log x = 36$ , $x$ is also equal to $b^{36}$ . Equating these, $b^{36} = 6^{108}$ $b = 6^3 = \boxed{216}$

2019 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2019 AIME II Problems/Problem 6

Problem 6

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