Difference between revisions of "1984 AIME Problems/Problem 5"

Revision as of 08:27, 18 October 2007

Problem

Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$ .

Solution

Use the change of base formula to see that $\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5$ ; combine denominators to find that $\frac{\log a^3b}{3\log 2} = 5$ . Doing the same thing with the second equation yields that $\frac{\log b^3a}{3\log 2} = 7$ . This means that $\log a^3b = 15\log 2 \Longrightarrow a^3b = 2^{15}$ and that $\log ab^3 = 21\log 2 \Longrightarrow ab^3 = 2^{21}$ . If we multiply the two equations together, we get that $a^4b^4 = 2^{36}$ , so taking the fourth root of that $ab = 2^9 = 512$ .

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 * [[American Invitational Mathematics Examination]]
 * [[Mathematics competition resources]]
+[[Category:Intermediate Algebra Problems]]

1984 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1984 AIME Problems/Problem 5"

Revision as of 08:27, 18 October 2007

Problem

Solution

See also