Simplify: (3)²(3)⁵ A) 30 B) 90 C) 59 049 D) 2187 E) 120

Question

Quizplus · Accepted Answer

By the simplification rules, we can expand the expression inside the parentheses that contains the number 3. 
So, we have: 
(3)S1U1P12S1S1P0(3)S1U1P15S1S1P0 = 3S1U1P12S1S1P0S1U1P15S1S1P0 
Next, we notice that the factors S1S1P0 and S1S1P0 are equal to the identity matrix I. 
So, we can simplify further: 
3S1U1P12S1U1P15 = 3S1U1P27 
Now, we apply the rule for exponentiation by squaring: 
S1U1P27 = (S1U1P9)³ 
S1U1P9 = (S1U1P3)³ 
S1U1P3 = (S1U1P1)³ 
S1U1P1 = S. 
Therefore, we obtain: 
3S1U1P12S1U1P15 = 3(S1U1P9)³ = 3(S1U1P3)⁹ = 3(S1U1P1)²⁷ = 3S²⁷. 
Finally, we apply the rule for exponentiation by squaring again: 
S²⁷ = (S¹³)² 
S¹³ = S(S⁴)⁴ = S(S(S²)²)⁴ = S(S(S)²)⁴ = S(S²)⁵ = Sⁱ²⁵. 
Therefore, 
3S¹³ = 3(S⁶)² = 3S⁷² = 3S²⁵⁶ = 2187.

Simplify: (3)2(3)5

Simplify: (3)²(3)⁵