Difference between revisions of "1983 AIME Problems/Problem 1"
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With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=\boxed{060}</math>. | With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=\boxed{060}</math>. | ||
− | ===Solution 2=== | + | === Solution 2 === |
First we'll convert everything to exponential form. | First we'll convert everything to exponential form. | ||
<math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. The only expression containing <math>z</math> is <math>(xyz)^{12}=w</math>. It now becomes clear that one way to find <math>\log_z w</math> is to find what <math>x^{12}</math> and <math>y^{12}</math> are in terms of <math>w</math>. | <math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. The only expression containing <math>z</math> is <math>(xyz)^{12}=w</math>. It now becomes clear that one way to find <math>\log_z w</math> is to find what <math>x^{12}</math> and <math>y^{12}</math> are in terms of <math>w</math>. |
Revision as of 15:01, 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,
.
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.