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

Revision as of 14:21, 6 May 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 $\log a^3b = 15\log 2 \Longrightarrow ab^3 = 2^{15}$ 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 $ab = 2^9 = 512$ .

@@ Line 9: / Line 9: @@
 {{AIME box|year=1984|num-b=4|num-a=6}}
+* [[AIME Problems and Solutions]]
+* [[American Invitational Mathematics Examination]]
+* [[Mathematics competition resources]]

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

Page

Toolbox

Search

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

Revision as of 14:21, 6 May 2007

Problem

Solution

See also