1983 AHSME Problems/Problem 25
Contents
Problem 25
If and , then is
Solution
We have that . We can substitute our value for 5, to get Hence Since , we have Therefore, we have
Solution 2
We have and . We can say that and .
We can evaluate (a+b) by the Addition Identity for Logarithms, . Also, .
Now we have to evaluate [2(1-b)], the denominator of the fractional exponent. Once again, we can say
~YBSuburbanTea
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.