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No, ( g ) does not equal ( g^g ) for most values of ( g ). The expression ( g^g ) represents ( g ) raised to the power of itself, which is typically greater than ( g ) when ( g > 1 ). For ( g = 1 ), both sides equal 1, and for ( g = 0 ), the expression ( g^g ) is defined as 1, while ( g ) is 0. Thus, the only case where ( g = g^g ) holds true is for ( g = 1 ).

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1mo ago

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