Q.1. (x+1)= 2*log(2^x+3)-2*log(1980-2^-x)first log has base 2 and second one has base 4. so, question is sum of roots of the above equation(a)3954 (b)log11(base=2) (c)log3954(base=2)Q.2. lim ((x^(x^x))-x^x) =?x tending to 0+

a.1)    option b

2 log(4) (1980 - 2^-x)    (4) means base  it can be written as 2/2 log(2) (1980 - 2^-x )  by log properties

putting the value 2^x = y finally we get the equation as

log {(y + 3 )}^2   - log {(1980 y -1)/y} = x + 1

logs are all base 2 now applying the log properties again we get

log [{y (y + 3)^2}/(1980y - 1)] = x + 1

y (y + 3)^2}/(1980y - 1) = 2 ^(x+1) = 2^x . 2

as 2^x = y we get

y^2 - 3954 y + 11 = 0

the roots will be in form of 2^x1 and 2^x2

there product   2^x1*2^x2 = 2^(x1+x2) = 11

x1+x2 = log (2) 11

q.2) lim x tending to 0+ x ^ x = 1

now putting this result in the limits and applying the properties of limits we get

-1 as answer

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