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

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=== Solution 1 ===
 
=== Solution 1 ===
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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 = \boxed{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 ab^3}{3\log 2} = 5</math>. Doing the same thing with the second equation yields that <math>\frac{\log a^3b}{3\log 2} = 7</math>. This means that <math>\log ab^3 = 15\log 2 \Longrightarrow ab^3 = 2^{15}</math> and that <math>\log a^3 b = 21\log 2 \Longrightarrow a^3 b = 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 = \boxed{512}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===

Revision as of 11:43, 12 June 2011

Problem

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

Solution 1

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 $\frac{\log ab^3}{3\log 2} = 5$. Doing the same thing with the second equation yields that $\frac{\log a^3b}{3\log 2} = 7$. This means that $\log ab^3 = 15\log 2 \Longrightarrow ab^3 = 2^{15}$ and that $\log a^3 b = 21\log 2 \Longrightarrow a^3 b = 2^{21}$. If we multiply the two equations together, we get that $a^4b^4 = 2^{36}$, so taking the fourth root of that, $ab = 2^9 = \boxed{512}$.

Solution 2

We can simplify our expressions by changing everything to a common base and by pulling exponents out of the logarithms. The given equations then become $\frac{\ln a}{\ln 8} + \frac{2 \ln b}{\ln 4} = 5$ and $\frac{\ln b}{\ln 8} + \frac{2 \ln a}{\ln 4} = 7$. Adding the equations and factoring, we get $(\frac{1}{\ln 8}+\frac{2}{\ln 4})(\ln a+ \ln b)=12$. Rearranging we see that $\ln ab = \frac{12}{\frac{1}{\ln 8}+\frac{2}{\ln 4}}$. Again, we pull exponents out of our logarithms to get $\ln ab = \frac{12}{\frac{1}{3 \ln 2} + \frac{2}{2 \ln 2}} = \frac{12 \ln 2}{\frac{1}{3} + 1} = \frac{12 \ln 2}{\frac{4}{3}} = 9 \ln 2$. This means that $\frac{\ln ab}{\ln 2} = 9$. The left-hand side can be interpreted as a base-2 logarithm, giving us $ab = 2^9 = \boxed{512}$.

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
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