Question

. scrieti sub forma unei singure puteri cu exponent numar natural:
a) 4 la puterea a 16 *8 la puterea 12*16 la puterea a 15=
b) 25 la puterea a 10*5 la puterea a 32*125 la puterea a 5=
c) 9 la puterea a 4*27 la puterea a 5*3 la puterea a 14=
d) a la puterea a 6*b la puterea a 2 *(a*b)la puterea a 4*b la puterea a 4=
e) (a*b) la puterea a 12:a la puterea a 4*b la puterea a 3*(a la puterea a 3*b)=
f) 36 la puterea a 5*6 la puterea a 5*6 la puterea x+1=


2. calculati:
a) [(120*625)*(2 la puterea a 3*10)]:(2*5 la puterea a 3:25)la puterea a 5=
b) 6+2*[93*5 la puterea a 2-2 la puterea a 2*18:3 la puterea a 2+3*11):2 la puterea a 2]=

Answer 1

Destul de laborioase, dar sa-i dam bice:

$4^{16}*8^{12}*16^{15} = (2^2)^{16}*(2^3)^{12}*(2^4)^{15} = 2^{32}*2^{36}*2^{60} = 2^{108}$
$25^{10}*5^{32}*125^{5}=(5^2)^{10}*5^{32}*(5^3)^{5}=5^{20}*5^{32}*5^{15}=5^{67}$
$9^4*27^5*3^{14} = (3^2)^4*(3^3)^5*3^{14} = 3^8*3^{15}*3^{14} = 3^{37}$
$a^6*b^2*(a*b)^4*b^4=a^{10}*b^{10} =(a*b)^{10}$
$(a*b)^{12}:a^4*b^3*(a^3*b)=a^{12}*b^{12}:a^4*b^3*a^3*b=a^{11}*b^b^{16}$
$36^5*6^{5}*6^{x+1} = (6^2)^5*6^5*6^{x+1}=6^{x+16}$