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

Latest revision as of 02:14, 17 August 2022

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 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 by multiplying the first two equations, we get, $x^{24}y^{40}= w^{2}$ . Taking the square root of this, we find that $x^{12}y^{20}=w$ . Recall, $x^{12}y^{12}z^{12}=w$ . Thus, $z^{12}= y^{8}$ . Also recall, $y^{40}=w$ . Therefore, $z^{60}$ = $y^{40}$ = $w$ . So, $\log_z w$ = $\boxed{060}$ .

-Dhillonr25, Bobbob

Solution 6

Converting all of the logarithms to exponentials gives $x^{24} = w, y^{40} =w,$ and $x^{12}y^{12}z^{12}=w.$ Thus, we have $y^{40} = x^{24} \Rightarrow z^3=y^2.$ We are looking for $\log_z w,$ which by substitution, is $\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.$

~coolmath2017

Video Solution

https://youtu.be/8XjBNtFWWww

~Lucas

@@ Line 1: / Line 1: @@
 == 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>\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_x w = 24</math>, <math>\log_y w = 40</math> and <math>\log_{xyz} w = 12</math>. Find <math>\log_z w</math>.
-__TOC__
+== Solution 1 ==
-== Solution ==
+The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.
-=== Solution 1 ===
-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>.
@@ Line 14: / Line 11: @@
 With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=\boxed{060}</math>.
-=== Solution 2 ===
+== 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 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^{\frac{1}{2}}</math>. Raising both sides of <math>y^{40}=w</math> to the <math>\frac{12}{40}</math>th power 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}w^{3/10}z^{12}=w</math>. Simplifying, we get <math>z^{60}=w</math>.
+So our answer is <math>\boxed{060}</math>.
+== Solution 3 ==
 Applying the change of base formula,
-<center><math>\begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\
+<cmath>\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*}</math></center>
+\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*}</cmath>
 Therefore, <math> \frac {\log z}{\log
 w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}</math>.
@@ Line 24: / Line 30: @@
 Hence, <math> \log_z w = \boxed{060}</math>.
+== Solution 4 ==
+Since <math>\log_a b = \frac{1}{\log_b a}</math>, the given conditions can be rewritten as <math>\log_w x = \frac{1}{24}</math>, <math>\log_w y = \frac{1}{40}</math>, and <math>\log_w xyz = \frac{1}{12}</math>. Since <math>\log_a \frac{b}{c} = \log_a b - \log_a c</math>,  <math>\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}</math>. Therefore, <math>\log_z w = \boxed{060}</math>.
+== 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 by multiplying the first two equations, we get, <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
+== Solution 6 ==
+Converting all of the logarithms to exponentials gives <math>x^{24} = w, y^{40} =w,</math> and <math>x^{12}y^{12}z^{12}=w.</math>
+Thus, we have <math>y^{40} = x^{24} \Rightarrow z^3=y^2.</math>
+We are looking for <math>\log_z w,</math> which by substitution, is <math>\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.</math>
+~coolmath2017
+== Video Solution ==
+https://youtu.be/8XjBNtFWWww
+~Lucas
 == See Also ==
@@ Line 29: / Line 54: @@
 [[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

1983 AIME (Problems • Answer Key • Resources)
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