Question

# If there are m and n numbers of positive integers for a and b respectively, satisfying 4(log10a)2+(log2b)2=1, then the value of m+n is

Solution

## 4(log10a)2+(log2b)2=1 ⇒(log2b)2=1−4(log10a)2 Since, LHS is non-negative, we should have 1−4(log10a)2≥0 ⇒(1−2log10a)(1+2log10a)≥0 ⇒−1≤2log10a≤1 ⇒−12≤log10a≤12 ⇒1√10≤a≤√10 Possible integers are 1,2,3 ∴m=3 Also, 4(log10a)2=1−(log2b)2 Since, LHS is non-negative, we should have 1−(log2b)2≥0 ⇒(1−log2b)(1+log2b)≥0 ⇒−1≤log2b≤1 ⇒12≤b≤2 Possible integers are 1,2 ∴n=2

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