Difference between revisions of "1983 AIME Problems/Problem 1"
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== Problem == | == Problem == | ||
− | 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_xw=24</math>, <math>\displaystyle \log_y w = 40</math>, and <math>\log_{xyz}w=12</math>. Find <math>\log_zw</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_xw=24</math>, <math>\displaystyle \log_y w = 40</math>, and <math>\log_{xyz}w=12</math>. Find <math>\log_zw</math>. |
== Solution == | == Solution == | ||
− | The | + | The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent [[exponential]] [[expression]]s. |
<math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. If we now convert everything to a power of <math>120</math>, it will be easy to isolate <math>z</math> and <math>w</math>. | <math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. If we now convert everything to a power of <math>120</math>, it will be easy to isolate <math>z</math> and <math>w</math>. |
Revision as of 12:17, 22 January 2007
Problem
Let ,
, and
all exceed
, and let
be a positive number such that
,
, and
. Find
.
Solution
The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential expressions.
,
, and
. If we now convert everything to a power of
, it will be easy to isolate
and
.
,
, and
.
With some substitution, we get and
.