Question 894861
logbb can be shown as log(b,b)
log2a can be shown as log(2,a)


your equation is log(2,a) + log(b,b) >= 6


since log(b,b) will always be equal to 1, then b can be any legitimate value greater than 0.


since log(b,b) will always be equal to 1, your equation becomes:


log(2,a) + 1 >= 6


subtract 1 from both sides of this equation to get log(2,a) >= 5


solve for log(2,a) = 5 to get log(2,a) = 5 if and only if 2^5 = a which occurs when a = 32.


so log(2,32) = 5


that appears to be the minimum value for a.
if a > 32, then log(2,a) will be greater than 5
if a < 32, then log(2,a) will be less than 5.


for example:


log(2,64) = 6 which is greater than 5.
log(2,16) = 4 which is less than 5.


a has to be greater than or equal to 32 and b has to be greater than or equal to 0.
therefore, a + b has to be greater than or equal to 32.


that's your solution as far as i can tell.