Antwort
Es ist egal, welcher Logarithmus Du benutzt!
3^(x+2) + 2^(x+2) + 2^x = 2^(x+5) + 3^x
2^(x+2) + 2^x - 2^(x+5) = 3^x - 3^(x+2)
2^x * (2^2 + 1 + 2^5) = 3^x * ( 1 - 3^2)
log ( 2^x * (2^2 + 1 + 2^5) ) = log ( 3^x * ( 1 - 3^2) )
log ( 2^x ) + log( 2^2 + 1 + 2^5 ) = log ( 3^x ) + log( 1 - 3^2 )
x * log ( 2 ) + log( 2^2 + 1 + 2^5 ) = x * log ( 3 ) + log( 1 - 3^2 )
x * log ( 2 ) - x * log ( 3 ) = log( 1 - 3^2 ) - log( 2^2 + 1 + 2^5 )
x * ( log ( 2 ) - log ( 3 ) ) = log( 1 - 3^2 ) - log( 2^2 + 1 + 2^5 )
x = ( log( 1 - 3^2 ) - log( 2^2 + 1 + 2^5 ) ) / ( ( log ( 2 ) - log ( 3 ) ) )