



 
Claire, log(baseb) a, written log_{b} a means the power to which you must raise b to get a. For example, log_{2} 8 = log_{2} 2^{3} = 3; i.e. you must raise 2 to the 3^{rd} power to get 8. Similarly, log_{2} 2 is 1; log_{2} (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 = log_{2} (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  


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