Difference between revisions of "2019 AIME II Problems/Problem 6"
Adam zheng (talk | contribs) (→Solution 7 (Easy System of Equations)) |
(→Solution 7 (Easy System of Equations)) |
||
(2 intermediate revisions by one other user not shown) | |||
Line 117: | Line 117: | ||
Solving this yields <math>m = 36</math>, <math>n = \frac{2}{3}</math>. | 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} | + | 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{216}</math>. |
~ adam_zheng | ~ adam_zheng | ||
+ | |||
+ | ==Solution 8 (Definition of Logarithm)== | ||
+ | The second equation implies that | ||
+ | <cmath>\log_{\log_b x} x=54\implies (\log_b x)^{54}=x \implies \log_b x=x^{\frac{1}{54}}</cmath> | ||
+ | The first equation implies that | ||
+ | <cmath>3\log_b(\sqrt{x} \log_b x)=56 \implies b^{\frac{56}{3}}=\sqrt{x} \log_b x</cmath> | ||
+ | Substituting the first result into the second gives us | ||
+ | <cmath>b^{\frac{56}{3}}=x^{\frac{1}{2}}\cdot x^{\frac{1}{54}} \implies b=x^{\frac{1}{36}}.</cmath> | ||
+ | Because <math>b^{36}=x,</math> <math>\log_b x=36</math> by the definition of a logarithm. | ||
+ | Substituting this into the second equation, | ||
+ | <cmath>\log_{36} x=54 \implies x=36^{54}.</cmath> | ||
+ | Finally, | ||
+ | <cmath>b=(36^{54})^{\frac{1}{36}}=36^{54\cdot\frac{1}{36}}=6^{2*\frac{3}{2}}=6^3=\boxed{216}.</cmath> | ||
==See Also== | ==See Also== |
Latest revision as of 10:09, 5 April 2024
Contents
Problem
In a Martian civilization, all logarithms whose bases are not specified are 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
Solution 7 (Easy System of Equations)
Using change of base on the second equation, we have
Using log rules on the first equation, we have
We notice that and
are in both equations. Thus, we set
and
and we have
Solving this yields ,
.
Now, , so we have
. Solving this yields
.
~ adam_zheng
Solution 8 (Definition of Logarithm)
The second equation implies that
The first equation implies that
Substituting the first result into the second gives us
Because
by the definition of a logarithm.
Substituting this into the second equation,
Finally,
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.