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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=14(log10a)2
Since, LHS is non-negative, we should have
14(log10a)20
(12log10a)(1+2log10a)0
12log10a1
12log10a12
110a10
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)20
(1log2b)(1+log2b)0
1log2b1
12b2
Possible integers are 1,2
n=2
 

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