# 2019 AIME II Problems/Problem 6

## Problem

In a Martian civilization, all logarithms whose bases are not specified are 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, $$\frac{\log x}{\log \log_{b}{x}}=54$$ $$\log x = 54 \cdot \log\log_{b}{x}$$ $$x = (\log_{b}{x})^{54}$$ Note by dolphin7 - you could also just rewrite the second equation in exponent form. Substituting this into the $\sqrt x$ of the first equation, $$3\log_{b}{((\log_{b}{x})^{27}\log_{b}{x})} = 56$$ $$3\log_{b}{(\log_{b}{x})^{28}} = 56$$ $$\log_{b}{(\log_{b}{x})^{84}} = 56$$

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

However, since we found that $\log_{b}{x} = 36$, $x$ 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 $$3\log_{b}{(\sqrt{x}\log x)}=\log_{b}{(x^{\frac{3}{2}}\log^3x)}=56$$ $$x^\frac{3}{2}\cdot \log_b^3x=b^{56}$$ 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

1st equation: $$\log (\sqrt{x}\log x)=\frac{56}{3}$$ $$\log(\sqrt x)+\log(\log x)=\frac{1}{2}\log x+\log(\log x)=\frac{56}{3}$$ 2nd equation: $$x=(\log x)^{54}$$ So now substitute $\log x=a$ and $x=b^a$: $$b^a=a^{54}$$ $$b=a^{\frac{54}{a}}$$ We also have that $$\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}$$ $$\frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}$$ This means that $\frac{14}{27}a=\frac{56}{3}$, so $$a=36$$ $$b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}$$.

-Stormersyle

## Solution 5 (Substitution)

Let $y = \log _{b} x$ Then we have $$3\log _{b} (y\sqrt{x}) = 56$$ $$\log _{y} x = 54$$ which gives $$y^{54} = x$$ Plugging this in gives $$3\log _{b} (y \cdot y^{27}) = 3\log _{b} y^{28} = 56$$ which gives $$\log _{b} y = \dfrac{2}{3}$$ so $$b^{2/3} = y$$ By substitution we have $$b^{36} = x$$ which gives $$y = \log _{b} x = 36$$ Plugging in again we get $$b = 36^{3/2} = \fbox{216}$$

--Hi3142

## Solution 6 (Also Substitution)

This system of equations looks complicated to work with, so we let $a=\log_bx$ to make it easier for us to read.

Now, the first equation becomes $3\log(\sqrt x \cdot a) = 56 \implies \log(\sqrt{x}\cdot a)=\frac{56}3$.

The second equation, $\log_{\log(x)}(x)=54$ gives us $\underline{a^{54} = x}$.

Let's plug this back into the first equation to see what we get: $\log_b(\sqrt{a^{54}}\cdot a)=\frac{56}3$, and simplifying, $\log_b(a^{27}\cdot a^1)=\log_b(a^{28})=\frac{56}{3}$, so $b^{\frac{56}3}=a^{28}\implies \underline{b^{\frac 23}=a}$.

Combining this new finding with what we had above $a^{54} = (b^{\frac 23})^{54} = x\implies \mathbf{b^{36} =x}$.

Now that we've expressed one variable in terms of the other, we can plug this into either equation, say equation 1. Then we get $\log_b(\sqrt{b^{36}}\cdot\log_b(b^{36})=\frac{56}3\implies$$\log_b(b^{18}\cdot 36)=\frac{56}3\implies b^{\frac{56}3}=b^{18}\cdot 36$.

Finally, that gives us that $\frac{b^{\frac{56}3}}{b^{18}}=36\implies b^{\frac{56}{3}-18}=b^{\frac{56}{3}-\frac{54}{3}}=b^{\frac 23}=36\implies b=36^{\frac 32}=6^3$. Thus, $b=\boxed{216}$.

~BakedPotato66

## Solution 7 (Easy System of Equations)

Using change of base on the second equation, we have

$$\frac{\log_{b} x}{\log_{b} \log_{b} x} = 54$$

Using log rules on the first equation, we have

$$\frac{3}{2} \log_{b} x + 3 \log_{b} \log_{b} x = 56$$

We notice that $\log_{b} x$ and $\log_{b} \log_{b} x$ are in both equations. Thus, we set $m = \log_{b} x$ and $n = \log_{b} \log_{b} x$ and we have

$$\frac{3}{2} m + 3n = 56$$ $$\frac{m}{n} = 54$$

Solving this yields $m = 36$, $n = \frac{2}{3}$.

Now, $n = \log_{b} \log_{b} x = \log_{b} m = \log_{b} 36$, so we have $\log_{b} 36 = \frac{2}{3}$. Solving this yields $b = \boxed{216}$.

The second equation implies that $$\log_{\log_b x} x=54\implies (\log_b x)^{54}=x \implies \log_b x=x^{\frac{1}{54}}$$ The first equation implies that $$3\log_b(\sqrt{x} \log_b x)=56 \implies b^{\frac{56}{3}}=\sqrt{x} \log_b x$$ Substituting the first result into the second gives us $$b^{\frac{56}{3}}=x^{\frac{1}{2}}\cdot x^{\frac{1}{54}} \implies b=x^{\frac{1}{36}}.$$ Because $b^{36}=x,$ $\log_b x=36$ by the definition of a logarithm. Substituting this into the second equation, $$\log_{36} x=54 \implies x=36^{54}.$$ Finally, $$b=(36^{54})^{\frac{1}{36}}=36^{54\cdot\frac{1}{36}}=6^{2*\frac{3}{2}}=6^3=\boxed{216}.$$