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, \[b^{\frac{2}{3}}=36\Rightarrow \boxed{b=216}\]


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 \[a=36\] \[b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}\].


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 \[b = 36^{3/2} = \fbox{216}\]


See Also

2019 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions

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