Difference between revisions of "1983 AIME Problems/Problem 1"
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== Problem == | == 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> | + | 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>. |
+ | __TOC__ | ||
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent [[exponential]] [[expression]]s. | The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent [[exponential]] [[expression]]s. | ||
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<math>x^{120}=w^5</math>, <math>y^{120}=w^3</math>, and <math>(xyz)^{120}=w^{10}</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= | + | With some substitution, we get <math>w^5w^3z^{120}=w^{10}</math> and <math>\log_zw=\boxed{060}</math>. |
− | + | === Solution 2 === | |
− | + | 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} \\ |
− | Applying | + | \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> |
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Therefore, <math> \frac {\log z}{\log | Therefore, <math> \frac {\log z}{\log | ||
w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}</math>. | w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}</math>. | ||
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Hence, <math> \log_z w = 60</math>. | Hence, <math> \log_z w = 60</math>. | ||
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== See also == | == See also == | ||
{{AIME box|year=1983|before=First Question|num-a=2}} | {{AIME box|year=1983|before=First Question|num-a=2}} | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Revision as of 10:33, 25 April 2008
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 expressions.
, , 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
Applying the change of base formula,
\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*}$ (Error compiling LaTeX. Unknown error_msg)Therefore, .
Hence, .
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 |