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

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Determine the value of <math>ab</math> if <math>\log_8a+\log_4b^2=5</math> and <math>\log_8b+\log_4a^2=7</math>.
 
Determine the value of <math>ab</math> if <math>\log_8a+\log_4b^2=5</math> and <math>\log_8b+\log_4a^2=7</math>.
  
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== Solution ==
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=== Solution 1 ===
 
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 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>.
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=== Solution 2 ===
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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 <math>\frac{\ln a}{\ln 8} + \frac{2 \ln b}{\ln 4} = 5</math> and <math>\frac{\ln b}{\ln 8} + \frac{2 \ln a}{\n 4} = 7</math>. Adding the equations and factoring, we get <math>(\frac{1}{\ln 8}+\frac{2}{\ln 4})(\ln a+ \ln b)=12</math>. Rearranging we see that <math>\ln ab = \frac{12}{\frac{1}{\ln 8}+\frac{2}{\ln 4}}</math>. Again, we pull exponents out of our logarithms to get <math>\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</math>. This means that <math>\frac{\ln ab}{\ln 2} = 9 \ln 2</math>. The left-hand side can be interpreted as a base-2 logarithm, giving us <math>ab = 2^9 = \boxed{512}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:44, 5 March 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 $\frac{\log a^3b}{3\log 2} = 5$. 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, $ab = 2^9 = \boxed{512}$.

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 $\frac{\ln b}{\ln 8} + \frac{2 \ln a}{\n 4} = 7$ (Error compiling LaTeX. Unknown error_msg). 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 $\frac{\ln ab}{\ln 2} = 9 \ln 2$. 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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