Difference between revisions of "1983 AIME Problems/Problem 1"
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=== Solution 4 === | === Solution 4 === | ||
Since <math>\log_a b = \frac{1}{\log_b a}</math>, the given conditions can be rewritten as <math>\log_w x = \frac{1}{24}</math>, <math>\log_w y = \frac{1}{40}</math>, and <math>\log_w xyz = \frac{1}{12}</math>. Since <math>\log_a \frac{b}{c} = \log_a b - \log_a c</math>, <math>\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}</math>. Therefore, <math>\log_z w = \boxed{060}</math>. | Since <math>\log_a b = \frac{1}{\log_b a}</math>, the given conditions can be rewritten as <math>\log_w x = \frac{1}{24}</math>, <math>\log_w y = \frac{1}{40}</math>, and <math>\log_w xyz = \frac{1}{12}</math>. Since <math>\log_a \frac{b}{c} = \log_a b - \log_a c</math>, <math>\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}</math>. Therefore, <math>\log_z w = \boxed{060}</math>. | ||
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+ | === Solution 5 === | ||
+ | |||
+ | If we convert all of the equations into exponential form, we receive <math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. The last equation can also be written as <math>x^{12}y^{12}z^{12}=w</math>. Also note that <math>x^{24}y^{40}=</math>w^2. Taking the square root of this, we find that <math>x^{12}y^{20}=w</math>. However, <math>x^{12}y^{12}z^{12}=w</math>. Thus, after <math>z^{12}=</math>y^{8}. Thus <math>\log_z w</math>= 12<math>{log_y w}/{8}</math>= \boxed{060}$. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 15:04, 8 December 2020
Contents
[hide]Problem
Let ,
and
all exceed
and let
be a positive number such that
,
and
. Find
.
Solution
Solution 1
The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.
,
, and
. If we now convert everything to a power of
, it will be easy to isolate
and
.
,
, and
.
With some substitution, we get and
.
Solution 2
First we'll convert everything to exponential form.
,
, and
. The only expression containing
is
. It now becomes clear that one way to find
is to find what
and
are in terms of
.
Taking the square root of the equation results in
. Raising both sides of
to the
th power gives
.
Going back to , we can substitute the
and
with
and
, respectively. We now have
. Simplifying, we get
.
So our answer is
.
Solution 3
Applying the change of base formula,
Therefore,
.
Hence, .
Solution 4
Since , the given conditions can be rewritten as
,
, and
. Since
,
. Therefore,
.
Solution 5
If we convert all of the equations into exponential form, we receive ,
, and
. The last equation can also be written as
. Also note that
w^2. Taking the square root of this, we find that
. However,
. Thus, after
y^{8}. Thus
= 12
= \boxed{060}$.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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.