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Shorten the following statements A=(2^5*2^-4)^2 *(3^3)^2 *3^-5

 Sep 19, 2015
 #1
avatar+130516 
+5

(2^5*2^-4)^2 *(3^3)^2 *3^-5

 

(2)^2 * (3)^6 * 3^(-5)  =

 

4 * 3  =

 

12

 

 

cool cool cool

 Sep 19, 2015
 #2
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0

=12

Here is how you get 12, step by step:

 

Simplify the following:
((2^5/2^4)^2 (3^3)^2)/3^5
For all positive integer exponents (a^n)^m = a^(n m). Apply this to (3^3)^2.
Multiply exponents. (3^3)^2 = 3^(3×2):
((2^5/2^4)^2 3^(3×2))/3^5
Multiply 3 and 2 together.
3×2  =  6:
((2^5/2^4)^2 3^6)/3^5
For all exponents, a^n/a^m = a^(n-m). Apply this to ((2^5/2^4)^2 3^6)/3^5.
Combine powers. ((2^5/2^4)^2 3^6)/3^5 = (2^5/2^4)^2 3^(6-5):
(2^5/2^4)^2 3^6-5
Subtract 5 from 6.
6-5 = 1:
(2^5/2^4)^2 3
For all exponents, a^n/a^m = a^(n-m). Apply this to 2^5/2^4.
Combine powers. 2^5/2^4 = 2^(5-4):
(2^5-4)^2 3
For all positive integer exponents (a^n)^m = a^(n m). Apply this to (2^(5-4))^2.
Multiply exponents. (2^(5-4))^2 = 2^((5-4)×2):
2^((5-4)×2)×3
Subtract 4 from 5.
5-4 = 1:
2^2×3
Evaluate 2^2.
2^2 = 4:
4×3
Multiply 4 and 3 together.
4×3  =  12:
Answer: | 
| 12

 Sep 19, 2015

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