Difference between revisions of "1983 AIME Problems/Problem 1"
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Let <math>x</math>, <math>y</math> and <math>z</math> all exceed <math>1</math> and let <math>w</math> be a positive number such that <math>\log_x w = 24</math>, <math>\log_y w = 40</math> and <math>\log_{xyz} w = 12</math>. Find <math>\log_z w</math>. | Let <math>x</math>, <math>y</math> and <math>z</math> all exceed <math>1</math> and let <math>w</math> be a positive number such that <math>\log_x w = 24</math>, <math>\log_y w = 40</math> and <math>\log_{xyz} w = 12</math>. Find <math>\log_z w</math>. | ||
− | == | + | == Solution == |
=== Solution 1 === | === Solution 1 === |
Revision as of 18:12, 15 February 2019
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
. Taking the
th root of
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, .
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.