1984 AIME Problems/Problem 5
Contents
Problem
Determine the value of if and .
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: Now add the two equations together then simplify, we'll get . So ,
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
Solution 6
Add the equations and use the facts that and to get Now use the change of base identity with base as 2: Which gives: Solving gives,
Solution 7
By properties of logarithms, we know that .
Using the fact that , we get .
Similarly, we know that .
From these two equations, we get and .
Multiply the two equations to get . Solving, we get that .
Solution 8
and adding both we get then we see is 3/2 times putting as x we get x+3x=12 so x=3 and = 3 so ab= = ~ math_comb01
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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