Question

Scrieti sub forma unei singure puteri cu exponent  numar natural:a)  4¹⁶ . 8¹².16¹⁵=
b)25¹°.5³².125⁵ =           c) 9⁴.27⁵.3¹⁴ =           d) a⁶.b².(a.b)⁴.b⁴ =      e)(a.b)¹²:a⁴.b³.(a².b) =
f) 36⁵.6⁵.6la puterea x+1 = 

Answer 1

a) 4¹⁶* 8¹²*16¹⁵=(2²)¹⁶ * (2³)¹² * (2⁴)¹⁵=2³² *2³⁶*2⁶°=2³²⁺³⁶⁺⁶°=2¹²⁸
b)25¹°*5³²*125⁵ = (5²)¹°*5³²*(5³)⁵=5²°*5³²*5¹⁵=5²°⁺³²⁺¹⁵=5⁶⁷
c) 9⁴*27⁵*3¹⁴ =(3²)⁴*(3³)⁵*3¹⁴=3⁸*3¹°*3¹⁴=3⁸⁺¹°⁺¹⁴=3³²
d) a⁶*b²*(a*b)⁴*b⁴ = a⁶*b²*a⁴*b⁴*b⁴=a⁶⁺⁴*b²⁺⁴⁺⁴=a¹°*b¹⁴
e)(a*b)¹²:a⁴*b³*(a²*b) =a¹²*b¹² : a⁴*b³*a²*b=a¹²⁻⁴⁺² * b¹²⁻³⁺¹=a¹° *b¹°
f) 36⁵*6⁵*$6^{x+1}$ = (6²)⁵ * 6⁵ * $6^{x+1}$=6¹° * 6⁵ *
$6^{x+1}$=$6^{10+5+x+1}$=$6^{16+x}$