Difference between revisions of "2019 AIME II Problems/Problem 6"

Revision as of 11:58, 6 January 2024

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 $b = \boxed{36}$ .

~ adam_zheng

@@ Line 117: / Line 117: @@
 Solving this yields <math>m = 36</math>, <math>n = \frac{2}{3}</math>.
-Now, <math>n = \log_{b} \log_{b} x = \log_{b} m = \log_{b} m</math>, so we have <math>\log_{b} 36 = \frac{2}{3}</math>. Solving this yields <math>b = \boxed{36}</math>.
+Now, <math>n = \log_{b} \log_{b} x = \log_{b} m = \log_{b} 36</math>, so we have <math>\log_{b} 36 = \frac{2}{3}</math>. Solving this yields <math>b = \boxed{36}</math>.
 ~ adam_zheng

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