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(1/9)^x=3*sqrt(27) Solve for x, without log
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Well, prime factorize both, I guess
\[ \frac{1}{9}=3^{-2} \]And\[ 3\sqrt{27} = 3^{1+\frac32} \]
So \[ -2x=\frac 52 \]I guess
\(\large\color{black}{ (1/9)^x=3\sqrt{27} }\) \(\large\color{black}{ (9)^{-x}=9\sqrt{3} }\) \(\large\color{black}{ (9)^{-x-1}=\sqrt{3} }\) \(\large\color{black}{ (9)^{-x-1}=9^{1/4}}\)
making sense?
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I am getting the same result, I think.
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