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 (Elegant)

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

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 $3\log _{b} (y * y^{27}) = 3\log _{b} y^{28} = 56$ 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

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