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

Revision as of 15:04, 8 December 2020

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 $\log_z w$ .

Solution

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 $x^{12}=w^{\frac{1}{2}}$ . Raising both sides of $y^{40}=w$ to the $\frac{12}{40}$ th power 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, $\log_z w = \boxed{060}$ .

Solution 4

Since $\log_a b = \frac{1}{\log_b a}$ , the given conditions can be rewritten as $\log_w x = \frac{1}{24}$ , $\log_w y = \frac{1}{40}$ , and $\log_w xyz = \frac{1}{12}$ . Since $\log_a \frac{b}{c} = \log_a b - \log_a c$ , $\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}$ . Therefore, $\log_z w = \boxed{060}$ .

Solution 5

If we convert all of the equations into exponential form, we receive $x^{24}=w$ , $y^{40}=w$ , and $(xyz)^{12}=w$ . The last equation can also be written as $x^{12}y^{12}z^{12}=w$ . Also note that $x^{24}y^{40}=$ w^2. Taking the square root of this, we find that $x^{12}y^{20}=w$ . However, $x^{12}y^{12}z^{12}=w$ . Thus, after $z^{12}=$ y^{8}. Thus $\log_z w$ = 12 ${log_y w}/{8}$ = \boxed{060}$.

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

@@ Line 37: / Line 37: @@
 === Solution 5 ===
-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}=</math>w^2. Taking the square root of this, we find that <math>x^{12}y^{20}=w</math>. However, <math>x^{12}y^{12}z^{12}=w</math>. Thus, after <math>z^{12}=</math>y^{8}. Thus <math>\log_z w</math>= 12<math>{log_y w}/{8}</math>= \boxed{060}$.
+If we convert all of the equations into exponential form, we receive <math>x^{24}=w</math>, <math>y^{40}=w</math>, and <math>(xyz)^{12}=w</math>. The last equation can also be written as <math>x^{12}y^{12}z^{12}=w</math>. Also note that <math>x^{24}y^{40}=</math>w^2. Taking the square root of this, we find that <math>x^{12}y^{20}=w</math>. However, <math>x^{12}y^{12}z^{12}=w</math>. Thus, after <math>z^{12}=</math>y^{8}. Thus <math>\log_z w</math>= 12<math>{log_y w}/{8}</math> = \boxed{060}$.
 {{alternate solutions}}

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