Difference between revisions of "1983 AIME Problems/Problem 1"
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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}= w^{2}</math>. Taking the square root of this, we find that <math>x^{12}y^{20}=w</math>. Recall, <math>x^{12}y^{12}z^{12}=w</math>. Thus, <math>z^{12}= y^{8}</math>. Also recall, <math>y^{40}=w</math>. Therefore, <math>z^{60}</math> = <math>y^{40}</math> = <math>w</math>. So, <math>\log_z w</math> = <math>\boxed{060}</math>. | 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}= w^{2}</math>. Taking the square root of this, we find that <math>x^{12}y^{20}=w</math>. Recall, <math>x^{12}y^{12}z^{12}=w</math>. Thus, <math>z^{12}= y^{8}</math>. Also recall, <math>y^{40}=w</math>. Therefore, <math>z^{60}</math> = <math>y^{40}</math> = <math>w</math>. So, <math>\log_z w</math> = <math>\boxed{060}</math>. | ||
+ | |||
+ | -Dhillonr25, Bobbob | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 16:10, 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
. Taking the square root of this, we find that
. Recall,
. Thus,
. Also recall,
. Therefore,
=
=
. So,
=
.
-Dhillonr25, Bobbob
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.