Question

14 si 18 vă rog, repede •u•

Answer 1

14. Se foloseste relatia demonstrata la ex 15 ( a^n=(a-1)[1+a+a^2+...+a^(n-1)] )

a=1+2^1+2^2+...+2^101=(2^102-1)÷(2-1)=2^102-1

b=3(1+4^1+4^2+...+4^50)=3(4^51-1)÷(4-1)=3(4^51-1)÷3=4^51-1=(2^2)^51-1=2^102-1=a

18. a) (4^x)^2=2^8 ⇔ [(2^2)^x]^2=2^8 ⇔ 2^(2·x·2)=2^8 ⇔ 2^(4·x)=2^8 ⇔4x=8 ⇔x=2