Claire,

log(baseb) a, written log_b a means the power to which you must raise b to get a.

For example, log₂ 8 = log₂ 2³ = 3; i.e. you must raise 2 to the 3^rd power to get 8.

Similarly, log₂ 2 is 1; log₂ (1/2) = -1 since 1/2 = 2^-1, etc.

Now what would log_(½) 2 be? It's the power to which you must raise 1/2 to get 2.

But (½)^-1 = 1/(½) = 2 thus log_(½) 2 = -1.

Note that we've just seen log_(½) 2 = log₂ (½) = -1 so that we have a solution to your problem of a = 2 and b = ½. I think you can see that there would be infinitely many solutions.

Penny