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

Revision as of 18:03, 15 February 2019

Problem

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

Solutions

Solution 1

The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.

$x^{24}=w$ , $y^{40}=w$ , 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 $\log_zw=\boxed{060}$ .

Solution 2

First we'll convert everything to exponential form. $x^{24}=w$ , $y^{40}=w$ , and $(xyz)^{12}=w$ . The only expression containing $z$ is $(xyz)^{12}=w$ . It now becomes clear that one way to find $\log_z w$ is to find what $x^{12}$ and $y^{12}$ are in terms of $w$ .

Taking the square root of the equation $x^{24}=w$ results in $x^{12}=w^{\frac{1}{2}}$ . Taking the $\frac{12}{40}$ th root of $y^{40}=w$ gives $y^{12}=w^{\frac{3}{10}}$ .

Going back to $(xyz)^{12}=w$ , we can substitute the $x^{12}$ and $y^{12}$ with $w^{1/2}$ and $w^{3/10}$ , respectively. We now have $w^{1/2}w^{3/10}z^{12}=w$ . Simplifying, we get $z^{60}=w$ . So our answer is $\boxed{060}$ .

Solution 3

Applying the change of base formula, $\begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\ \log_y w = 40 &\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\ \log_{xyz} w = 12 &\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*}$ Therefore, $\frac {\log z}{\log w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}$ .

Hence, $\log_z w = \boxed{060}$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 15: / Line 15: @@
 ===Solution 2===
 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 with z is <math>(xyz)^{12}=w</math>. It now becomes clear 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>.
-Taking the square root of the equation <math>x^{24}=w</math> results in <math>x^{12}=w^{1/2}</math>. Taking the <math>12/40</math> root of <math>y^{40}=w</math> equates to <math>y^{12}=w^{3/10}</math>.
+Taking the square root of the equation <math>x^{24}=w</math> results in <math>x^{12}=w^{\frac{1}{2}}</math>. Taking the <math>\frac{12}{40}</math>th root of <math>y^{40}=w</math> gives <math>y^{12}=w^{\frac{3}{10}}</math>.
-Going back to <math>(xyz)^{12}=w</math>, we can substitute the <math>x^{12}</math> and <math>y^{12}</math> with <math>w^{1/2}</math> and <math>w^{3/10}</math>, respectively. We now have <math>w^{1/2}</math>*<math>w^{3/10}</math>*<math>z^{12}=w</math>. Simplify we get <math>z^{60}=w</math>.
+Going back to <math>(xyz)^{12}=w</math>, we can substitute the <math>x^{12}</math> and <math>y^{12}</math> with <math>w^{1/2}</math> and <math>w^{3/10}</math>, respectively. We now have <math>w^{1/2}w^{3/10}z^{12}=w</math>. Simplifying, we get <math>z^{60}=w</math>.
 So our answer is <math>\boxed{060}</math>.

1983 AIME (Problems • Answer Key • Resources)
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