Matematyka

Autorzy:Aleksandra Ciszkowska, Alina Przychoda, Zygmunt Łaszczyk

Wydawnictwo:WSiP

Rok wydania:2012

Oblicz: a) log_(1/5) | log_3 1/243| 4.5 gwiazdek na podstawie 8 opinii
  1. Liceum
  2. 1 Klasa
  3. Matematyka

a)

`log_(1/5) |log_3 1/243|=log_(1/5) | log_3 1/3^5|=log_(1/5) | log_3 3^(-5)|=`

`=log_(1/5) |(-5)|=log_(1/5) 5=log_(1/5) (1/5)^(-1)= ul(ul(-1))`

b)

`log_5(5(log800+log125))=log_5(5*log100000)=`

`=log_5 (5*log 10^5)=log_5 (5*5)=log_5 5^2=ul(ul2)`

c)

Skorzystamy ze wzoru na zamianę podstawy logarytmu:

`log_a b= (log_x b)/(log_x a)`

 

`((sqrt3)^(1/2log_3 5) +(log_3 5 * log_5 243)^(1/2))(root(8)25-(root(16)5)^8)=`

`=((3^(1/2))^(1/2log_3 5) +(strike(log_3 5) * (log_3 243)/strike(log_3 5))^(1/2))(25^(1/8)-(5^(1/strike16))^strike8)=`    

`=(3^(1/4log_3 5) +(log_3 243)^(1/2))((5^strike2)^(1/strike8)-(5^(1/2)))=`   

`=(3^(log_3 5^(1/4)) +(log_3 3^5)^(1/2))(5^(1/4)-5^(1/2))=`   

`=(5^(1/4) +5^(1/2))(5^(1/4)-5^(1/2)) \ \ \ stackrel((a+b)(a-b)=a^2-b^2)= \ \ \ ` `(5^(1/strike4))^strike2-(5^(1/strike2))^strike2=`  

    

`=5^(1/2)-5^1=ul(ul(sqrt5-5))`  

d)

`(5log_21 21- log_21 441+ log_21 1)^(log_9 49)=` `(5*1-log_21 21^2+ log_21 21^0)^(log_9 49)=` 

`=(5-2+0)^(log_9 49)=` `3^(log_9 49)= (sqrt9)^(log_9 49)= (9^(1/2))^(log_9 49)=`

`=9^(1/2log_9 49)= 9^(log_9 49^(1/2))=49^(1/2)=sqrt49=ul(ul7)`