Math hw help please
11-26-2007, 09:06 PM
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Math hw help please Ok im stuck on this one can any math people help me please. Thanks
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11-26-2007, 10:33 PM
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This is true but don't forget that:
log9(1/x) = log9(1) -log9(x)
noting:
log9(1) = 0
therefore:
-3 = -log9(x)
or:
3 = log9(x)
Answer E.
Or a simpler method would be to say that:
9^-3 = 1/x = x^-1
so:
9^3 = x^1 = x
or:
3 = log9(x)
Answer E.
Somebody should check my sanity but I believe that is the correct answer.
11-26-2007, 11:27 PM
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This is true but don't forget that:
log9(1/x) = log9(1) -log9(x)
noting:
log9(1) = 0
therefore:
-3 = -log9(x)
or:
3 = log9(x)
Answer E.
Or a simpler method would be to say that:
9^-3 = 1/x = x^-1
so:
9^3 = x^1 = x
or:
3 = log9(x)
Answer E.
Somebody should check my sanity but I believe that is the correct answer.
log9(1/x) = log9(1) -log9(x)
noting:
log9(1) = 0
therefore:
-3 = -log9(x)
or:
3 = log9(x)
Answer E.
Or a simpler method would be to say that:
9^-3 = 1/x = x^-1
so:
9^3 = x^1 = x
or:
3 = log9(x)
Answer E.
Somebody should check my sanity but I believe that is the correct answer.
