Name: sandy

Question: find X 3/2logb4-1/2logb8+logb2=logbx

my book uses prop 7 to rewrite as logb4raised3/2-logb8raised2/3+logb2=logbx; then writes logb8-logb4+logb2=logbx how did it get here?

Hi Sandy,

I get the second part

log_{b}4^{3/2} = log_{b}8

because

4^{3/2} = 2^{(2)*(3/2)} = 2^{3} = 8

and

log_{b}8^{2/3} = log_{b}4

8^{2/3} = 2^{(3)*(2/3)} = 2^{2} = 4.

I don't know how you get log_{b}8^{2/3} from 1/2log_{b}8 though. I hope the 1/2 was a typo that should be a 2/3.

Claude

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