Difference between revisions of "1983 AIME Problems/Problem 1"
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== Problem == | == Problem == | ||
− | Let <math>x</math>, <math>y</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>. |
− | == | + | == Solution == |
=== Solution 1 === | === Solution 1 === | ||
− | The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential | + | The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms. |
<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>. | ||
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With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=\boxed{060}</math>. | 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. | 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 | + | <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 | + | 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} | + | 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>. | So our answer is <math>\boxed{060}</math>. | ||
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Hence, <math> \log_z w = \boxed{060}</math>. | 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 3,</math> which by substitution, is <math>\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.</math> | ||
+ | |||
+ | ~coolmath2017 | ||
== See Also == | == See Also == |
Latest revision as of 16:38, 18 December 2020
Contents
Problem
Let , and all exceed and let be a positive number such that , and . Find .
Solution
Solution 1
The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.
, , and . If we now convert everything to a power of , it will be easy to isolate and .
, , and .
With some substitution, we get and .
Solution 2
First we'll convert everything to exponential form. , , and . The only expression containing is . It now becomes clear that one way to find is to find what and are in terms of .
Taking the square root of the equation results in . Raising both sides of to the th power gives .
Going back to , we can substitute the and with and , respectively. We now have . Simplifying, we get . So our answer is .
Solution 3
Applying the change of base formula, Therefore, .
Hence, .
Solution 4
Since , the given conditions can be rewritten as , , and . Since , . Therefore, .
Solution 5
If we convert all of the equations into exponential form, we receive , , and . The last equation can also be written as . Also note that by multiplying the first two equations, we get, . Taking the square root of this, we find that . Recall, . Thus, . Also recall, . Therefore, = = . So, = .
-Dhillonr25, Bobbob
Solution 6
Converting all of the logarithms to exponentials gives and Thus, we have We are looking for which by substitution, is
~coolmath2017
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.