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2^n=radical(3^(n-2))
Guest Feb 10, 2012
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2^n=sqrt(3^(n-2))

:: if nothing else seems to work, try 'ln( )' on both sides:
ln( 2^n ) = ln( sqrt(3^(n-2)) )
:: logarithm law: ln( x^y ) => y*ln(x) (x,y >0)
:: sqrt( x ) = x^(1/2)
n*ln(2) = ln( (3^(n-2))^(1/2) )
:: pow law
n*ln(2) = ln( (3^(1/2*(n-2))) )
n*ln(2) = 1/2*(n-2)*ln( 3 )
n*ln(2) = 1/2*n*ln( 3 ) - ln( 3 )
n*ln(2) - 1/2*n*ln( 3 ) = -ln( 3 )
n*( ln(2) - 1/2*ln(3) ) = -ln( 3 )
n = -ln( 3 ) / ( ln(2) - 1/2*ln(3) )
n = -2*ln( 3 ) / ( 2*ln(2) - ln(3) )
admin  Feb 11, 2012

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