Difference between revisions of "1984 AIME Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | Use the [[change of base formula]] to see that <math>\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5</math>; combine [[denominator]]s to find that <math>\frac{\log a^3b}{3\log 2} = 5</math>. Doing the same thing with the second equation yields that <math>\frac{\log b^3a}{3\log 2} = 7</math>. This means that <math>\log a^3b = 15\log 2 \Longrightarrow a^3b = 2^{15}</math> and that <math>\log ab^3 = 21\log 2 \Longrightarrow ab^3 = 2^{21}</math>. If we multiply the two equations together, we get that <math>a^4b^4 = 2^{36}</math>, so taking the fourth root of that <math>ab = 2^9 = 512</math>. | + | Use the [[change of base formula]] to see that <math>\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5</math>; combine [[denominator]]s to find that <math>\frac{\log a^3b}{3\log 2} = 5</math>. Doing the same thing with the second equation yields that <math>\frac{\log b^3a}{3\log 2} = 7</math>. This means that <math>\log a^3b = 15\log 2 \Longrightarrow a^3b = 2^{15}</math> and that <math>\log ab^3 = 21\log 2 \Longrightarrow ab^3 = 2^{21}</math>. If we multiply the two equations together, we get that <math>a^4b^4 = 2^{36}</math>, so taking the fourth root of that, <math>ab = 2^9 = 512</math>. |
== See also == | == See also == | ||
Revision as of 11:50, 18 October 2007
Problem
Determine the value of 
if 
and 
.
Solution
Use the change of base formula to see that 
; combine denominators to find that 
. Doing the same thing with the second equation yields that 
. This means that 
and that 
. If we multiply the two equations together, we get that 
, so taking the fourth root of that, 
.
See also
| 1984 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
