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log₃(logx³ )=5⁰
log₃(logx³ )=1
logx³ =3¹
logx³ =3
3logx=3
logx=1
x=10nalezy spr. zał, tzn, że liczba logarytmowana >0
ale pasuje
b) X= log₈[log₄(log₃9)]
X= log₈[log₄(log₃3²)]= log₈[log₄(2)]= log₈[log₄(4)^ 1/2]=
log₈[ 1/2]= log₈ 1 - log₈ 2= 0 - log₈ 2 = - log₈ 2 = - log₈ 8^ 1/3 = -1/3
dziedzina
x>0
logx³>0 czyli x³>1
log₃(logx³ )>0, czyli logx³>1, czyli x³>10
5⁰=log₃(logx³ )
1=log₃(logx³ )
3¹=logx³
3=logx³
10³=x³
x=10 nalezy do dziedziny
b) x= log₈[log₄(log₃9)]
x=log₈(log₄2)
x=log₈(½)
x=-3