Difference between revisions of "1984 AIME Problems/Problem 5"
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=== Solution 4 === | === Solution 4 === | ||
We can change everything to a common base, like so: <math>\log_8{a} + \log_8{b^3} = 5,</math> <math>\log_8{b} + \log_8{a^3} = 7.</math> We set the value of <math>\log_8{a}</math> to <math>x</math>, and the value of <math>\log_8{b}</math> to <math>y.</math> Now we have a system of linear equations: <cmath>x + 3y = 5,</cmath> <cmath>y + 3x = 7.</cmath> We multiply the second equation by three and subtract the first equation from the second to get <math>8x = 16,</math> which gives <math>x=2.</math> We substitute our value into the first equation and find that <math>y = 1.</math> Now we know that <math>\log_8{a} = 2,</math> and <math>\log_8{b} = 1,</math> so we find that <math>a = 64,</math> and <math>b=8.</math> Multiplying together gives us <math>ab = \boxed{512}.</math> | We can change everything to a common base, like so: <math>\log_8{a} + \log_8{b^3} = 5,</math> <math>\log_8{b} + \log_8{a^3} = 7.</math> We set the value of <math>\log_8{a}</math> to <math>x</math>, and the value of <math>\log_8{b}</math> to <math>y.</math> Now we have a system of linear equations: <cmath>x + 3y = 5,</cmath> <cmath>y + 3x = 7.</cmath> We multiply the second equation by three and subtract the first equation from the second to get <math>8x = 16,</math> which gives <math>x=2.</math> We substitute our value into the first equation and find that <math>y = 1.</math> Now we know that <math>\log_8{a} = 2,</math> and <math>\log_8{b} = 1,</math> so we find that <math>a = 64,</math> and <math>b=8.</math> Multiplying together gives us <math>ab = \boxed{512}.</math> | ||
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+ | === Solution 5 === | ||
+ | Add the two equations to get <math>\log_8 {a}+ \log_8 {b}+ \log_{a^2}+\log_{b^2}=12</math>. This can be simplified with the log property <math>\log_n {x}+\log_n {y}=log_n {xy}</math>. Using this, we get <math>\log_8 {ab}+ \log_4 {a^2b^2}=12</math>. Now let <math>\log_8 {ab}=c</math> and <math>\log_4 {a^2b^2}=k</math>. Converting to exponents, we get <math>8^c=ab</math> and <math>4^k=(ab)^2</math>. Sub in the <math>8^c</math> to get <math>k=3c</math>. So now we have that <math>k+c=12</math> and <math>k=3c</math> which gives <math>c=3</math>, <math>k=9</math>. This means <math>\log_4 {a^2b^2}=9</math> so <math>4^9=(ab)^2 \implies ab=(2^2)^9 \implies 2^9 \implies \boxed {512}</math> | ||
== See also == | == See also == |
Revision as of 20:57, 23 September 2016
Contents
Problem
Determine the value of if and .
Solution
Solution 1
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, .
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 and . Adding the equations and factoring, we get . Rearranging we see that . Again, we pull exponents out of our logarithms to get . This means that . The left-hand side can be interpreted as a base-2 logarithm, giving us .
Solution 3
This solution is very similar to the above two, but it utilizes the well-known fact that Thus, Similarly, Adding these two equations, we have .
Solution 4
We can change everything to a common base, like so: We set the value of to , and the value of to Now we have a system of linear equations: We multiply the second equation by three and subtract the first equation from the second to get which gives We substitute our value into the first equation and find that Now we know that and so we find that and Multiplying together gives us
Solution 5
Add the two equations to get . This can be simplified with the log property . Using this, we get . Now let and . Converting to exponents, we get and . Sub in the to get . So now we have that and which gives , . This means so
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 |