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OpenStudy (anonymous):
Can someone explain how to solve this? 3^x^3 = 3^3^x * 1/3^2
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OpenStudy (anonymous):
\[3^{x ^{3}}=3^{3x}*3^{-2}\] \[3^{x ^{3}}=3^{3x-2}\] x^3=3x-2 x^3-3x+2=0 (x-1)^2*(x+2)=0 so x=1 or x=-2
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