Respuesta :
FUN Q!
look for 1/1000 < x < 1000 where x=2^n, n must be an integer
taking log on the inequalities
log1/1000 < logx < log1000
-3 < logx < 3
take log on x=2^n
logx=log (2^n)=nlog2=0.301n
substituting
-3 < 0.301n < 3
-9.9658 < n < 9.9658
n must be an integer: -9, -8,....0, 1,....9
ans is 19
u can repeat the same with log(3)
look for 1/1000 < x < 1000 where x=2^n, n must be an integer
taking log on the inequalities
log1/1000 < logx < log1000
-3 < logx < 3
take log on x=2^n
logx=log (2^n)=nlog2=0.301n
substituting
-3 < 0.301n < 3
-9.9658 < n < 9.9658
n must be an integer: -9, -8,....0, 1,....9
ans is 19
u can repeat the same with log(3)
following same reasoning as the above ans, log(1/1000) < log(3^n) < log(1000)
-3 < n*log3 < 3
-3 < 0.48n < 3
-6.3 < n < 6.3
n can be -6,-5,....5,6
total of 13 possible numbers for 3^n
