Difference between revisions of "2019 AIME II Problems/Problem 6"
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--Hi3142 | --Hi3142 | ||
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+ | ==Solution 6 (Also Substitution)== | ||
+ | This system of equations looks complicated to work with, so we let <math>a=\log_bx</math> to make it easier for us to read. | ||
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+ | Now, the first equation becomes <math>3\log(\sqrt x \cdot a) = 56 \implies \log(\sqrt{x}\cdot a)=\frac{56}3</math>. | ||
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+ | The second equation, <math>\log_{\log(x)}(x)=54</math> gives us <math>\underline{a^{54} = x}</math>. | ||
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+ | Let's plug this back into the first equation to see what we get: <math>\log_b(\sqrt{a^{54}}\cdot a)=\frac{56}3</math>, and simplifying, <math>\log_b(a^{27}\cdot a^1)=\log_b(a^{28})=\frac{56}{3}</math>, so <math>b^{\frac{56}3}=a^{28}\implies \underline{b^{\frac 23}=a}</math>. | ||
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+ | Combining this new finding with what we had above <math>a^{54} = (b^{\frac 23})^{54} = x\implies \mathbf{b^{36} =x}</math>. | ||
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+ | 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 <math>\log_b(\sqrt{b^{36}}\cdot\log_b(b^{36})=\frac{56}3\implies </math><math>\log_b(b^{18}\cdot 36)=\frac{56}3\implies b^{\frac{56}3}=b^{18}\cdot 36</math>. | ||
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+ | Finally, that gives us that <math>\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</math>. Thus, <math>b=\boxed{216}</math>. | ||
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+ | ~BakedPotato66 | ||
==See Also== | ==See Also== |
Revision as of 12:26, 5 February 2023
Contents
[hide]Problem
In a Martian civilization, all logarithms whose bases are not specified as assumed to be base , for some fixed . A Martian student writes down and finds that this system of equations has a single real number solution . Find .
Solution 1
Using change of base on the second equation to base b, Note by dolphin7 - you could also just rewrite the second equation in exponent form. Substituting this into the of the first equation,
We can manipulate this equation to be able to substitute a couple more times:
However, since we found that , is also equal to . Equating these,
Solution 2
We start by simplifying the first equation to Next, we simplify the second equation to Substituting this into the first equation gives Plugging this into gives -ktong
Solution 3
Apply change of base to to yield: which can be rearranged as: Apply log properties to to yield: Substituting into the equation yields: So Substituting this back in to yields So,
-Ghazt2002
Solution 4
1st equation: 2nd equation: So now substitute and : We also have that This means that , so .
-Stormersyle
Solution 5 (Substitution)
Let Then we have which gives Plugging this in gives which gives so By substitution we have which gives Plugging in again we get
--Hi3142
Solution 6 (Also Substitution)
This system of equations looks complicated to work with, so we let to make it easier for us to read.
Now, the first equation becomes .
The second equation, gives us .
Let's plug this back into the first equation to see what we get: , and simplifying, , so .
Combining this new finding with what we had above .
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 .
Finally, that gives us that . Thus, .
~BakedPotato66
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.