# 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 1

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, $$b^{36} = 6^{108}$$ $$b = 6^3 = \boxed{216}$$

## Solution 2

We start by simplifying the first equation to $$3\log(\sqrt{x}\log x)=\log(x^{\frac{3}{2}}\log^3x)=56$$ $$x^\frac{3}{2}\cdot \log_b^3x=b^{56}$$ Next, we simplify the second equation to $$\log_{\log(x)}(x)=\frac{\log_b(x)}{\log_b(\log_b(x))}=54$$ $$\log_bx=54\log_b(\log_b(x))=\log_b(\log_b^{54}(x))$$ $$x=\log_b^{54}x$$ Substituting this into the first equation gives $$\log_b^{54\cdot \frac{3}{2}}(x)\cdot \log_b^3x=\log_b^{84}x=b^{56}$$ $$x=b^{b^{\frac{56}{84}}}=b^{b^{\frac{2}{3}}}$$ Plugging this into $x=\log_b^{54}x$ gives $$b^{b^{\frac{2}{3}}}=\log_b^{54}(b^{b^\frac{2}{3}})=b^{\frac{2}{3}\cdot 54}=b^{36}$$ $$b^{\frac{2}{3}}=36$$ $$b=36^{\frac{3}{2}}=6^3=\boxed{216}$$ -ktong

## Solution 3

Apply change of base to $$\log_{\log x}(x)=54$$ to yield: $$\frac{\log_b(x)}{\log_b(\log_b(x))}=54$$ which can be rearranged as: $$\frac{\log_b(x)}{54}=\log_b(\log_b(x))$$ Apply log properties to $$3\log(\sqrt{x}\log x)=56$$ to yield: $$3(\frac{1}{2}\log_b(x)+\log_b(\log_b(x)))=56\Rightarrow\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}$$ Substituting $$\frac{\log_b(x)}{54}=\log_b(\log_b(x))$$ into the equation $\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}$ yields: $$\frac{1}{2}\log_b(x)+\frac{\log_b(x)}{54}=\frac{28\log_b(x)}{54}=\frac{56}{3}$$ So $$\log_b(x)=36.$$ Substituting this back in to $$\frac{\log_b(x)}{54}=\log_b(\log_b(x))$$ yields $$\frac{36}{54}=\log_b(36).$$ So, $$b^{\frac{2}{3}}=36\Rightarrow \boxed{b=216}$$

-Ghazt2002

## Solution 4 (Easiest)

From the first equation we have that $\log (\sqrt{x}\log x)=\frac{56}{3}$, so $\log(\sqrt x)+\log(\log x)=\frac{1}{2}\log x+\log(\log x)=\frac{56}{3}$. From the second equation we have that $x=(\log x)^{54}$, so now set $\log x=a$ and $x=b^a$. Substituting, we have that $b^a=a^{54}$, so $b=a^{\frac{54}{a}}$. We also have that $\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}$, so $\frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}$. This means that $\frac{14}{27}a=\frac{56}{3}$, so $a=36$, and $b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}$.

-Stormersyle

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 