Problem:
For how many integers n does the expression
logn−3log(n2)−(logn)2
represent a real number, where log denotes the base 10 logarithm?
Answer Choices:
A. 900
B. 2
C. 902
D. 2
E. 901
Solution:
First note that the fraction under the square root will equal 0 when n=1. There are now 2 cases to consider: Either logn>3 or logn<3. In the first case, logn>3 gives n>1000, so the numerator log(n2)−(logn)2=2logn−(logn)2<0 because 2logn<(logn)2 whenever logn>2. Thus the fraction under the square root will be negative and there are no solutions in this case.
For the second case, logn<3, so n<1000. The fraction under the square root will be nonnegative provided log(n2)−(logn)2≤0. This will occur when logn≥2, i.e., when n≥100. Thus the given expression will evaluate to a real number for any integer n in the range 100≤n<1000. Together with the n=1 solution, this gives a total of (E)901 integers for which the expression evaluates to a real number.
OR
If the expression represents a real number, then logn is defined, so n>0. Because n is an integer, n>0 implies n≥1. Let m=logn≥log1=0. The fraction under the square root can be expressed as
m−32m−m2=m−3m(2−m)
This expression is meaningful and nonnegative when m=0 or 2≤m<3, as shown in the table below. If m=0, then n=1. When 2≤m<3, it follows that 100≤n<1000 and hence another 900 integral values of n make the expression represent a real number. This gives a total of 1+900=(E)901 possible values of n.