Difference between revisions of "1983 AIME Problems/Problem 1"

Revision as of 22:49, 23 July 2006

Problem

Let $x$ , $y$ , and $z$ all exceed $1$ , and let $w$ be a positive number such that $\log_xw=24$ , $\displaystyle \log_y w = 40$ , and $\log_{xyz}w=12$ . Find $\log_zw$ .

Solution

The logarithmic notation doesn't tell us much, so we'll first convert everything to exponents.

$x^{24}=w$ , $y^{40}=w$ , and $(xyz)^{12}=w$ . If we now convert everything to a power of $120$ , it will be easy to isolate $z$ and $w$ .

$x^{120}=w^5$ , $y^{120}=w^3$ , and $(xyz)^{120}=w^{10}$ .

With some substitution, we get $w^5w^3z^{120}=w^{10}$ and $\log_zw=60$ .

@@ Line 3: / Line 3: @@
 == Solution ==
+The logarithmic notation doesn't tell us much, so we'll first convert everything to exponents.
+<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^{120}=w^5</math>, <math>y^{120}=w^3</math>, and <math>(xyz)^{120}=w^{10}</math>.
+With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=60</math>.
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Difference between revisions of "1983 AIME Problems/Problem 1"

Revision as of 22:49, 23 July 2006

Problem

Solution