1983 AHSME Problems/Problem 25
Contents
[hide]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 | |
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