Question

prosze prosze prosze o pomoc.

LOGARYTMY

oblicz wartos wyrazenia :

a)log(2)48 - log(2)3 =

[czyli logarytm z 48 przy podstawie z dwoch ]

b) log(3)1/6 + log(3)2/3 =

c) 2log(1/2)√3 + log(1/2)5 1/3 =

d) ( log(5)16 - log(5)80 ) doo potęgi 2 =

e) log(2/3)4 - 2log(2/3)3 =

f) log(√2)50 - log(√2)25 =

g) log(1/3)4 + log(1/3)6 - log(1/3)8 =

h) 2log(1/4)8 - 3log(√3)9 =

Answer 1

a)log₂48-log₂3=log₂(48/3)=log₂16=4

 

b)log₃⅙+log₃⅔=log₃(⅙×⅔)=log₃(2/18)=log₃ (1/9)=-2

 

c)2log(₁/₂) √3+log(₁/₂) 5⅓=2log(₁/₂)√3+log(₁/₂)5⅓=2log(₁/₂)√3+log(₁/₂ )16/3=

 

=log(₁/₂)√3²+log(₁/₂)16/3=log(₁/2)3+log₁/₂ 16/3=log₁/₂(3×16/3)=log₁/₂ 48/3=

 

=log₁/₂16=-4

 

d)log₅16-log₅80=log₅80/16=log₅5=1

 

e)log₂/₃4-2log₂/₃ 3=log₂/₃ 4-log₂/₃ 3²=log₂/₃ 4-log₂/₃ 9=log₂/₃ 4/9=log₂/₃ 2/3²=2

 

f)log(√₂)50-log(√₂)25=log(√₂) 50/25=log(√₂)2=1/2

 

g)log₁/₃)4+log₁/₃)6-log₁/₃)(4×6)-log₁/₃)8=log₁/₃)24-log₁/₃)8=log₁/₃) 24/8=log₁/₃)3=-1

 

h)2log(₁/₄)8-3log(√₃)9=log(₁/₄)8²-log√₃)9³=log₁/₄ 64-log√₃ 729=-4-6=-10