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\(\Large\log_{9}{\frac{1}{27}}\) I know how i would solve this using the change of base formula, but I know there's another way i can't remember. what is it?
write 27 as the exponent of 3 ...but late you'll need change of base formula...
*later
I'm...confused. There's no 3 in the equation
27 = 3^3 9 = 3^2
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\(\Large 9^x=\frac{1}{27}\rightarrow (3^2)^x=\frac{1}{3^3}\) ???
yes recall \(\Large \dfrac{1}{a^b} = a^{-b}\)
okay, thanks. i got it now
-3/2
yes :)
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