Prove the divisbility of the following numbers.

(12 to the power of 8) x (9 to the power of 12) is divisible by 6 to the power of 16

Answer: Blank x 6 to the power of 16

Answer should be an exponent with a base

3^16

Step-by-step explanation:

12^8 * 9^12

Rewrite 12 as 3*4 and 9 as 3*3

(3*4) ^8 * (3*3) ^12

We know that (ab) ^c = a^c * b^c

3^8 4^8 3^12 3^12

We can write 4 as 2^2

3^8 2^2^8 3^12 3^12

We know a^b^c = a^ (b*c)

3^8 2^ (2*8) 3^12 3^12

3^8 2^ (16) 3^12 3^12

We also know that a^b * a^c * a^d = a^ (b+c+d)

2^ (16) 3^8 3^12 3^12

2^ (16) 3^ (8+12+12)

2^16 3^ (32)

But we need 6 ^16 so we will need a 3^16 3^32 = 3^16 * ^16 (16+16=32)

2^16 * 3^16 * 3^16

Remember a^b*c^b = (ac) ^b

(2*3) ^16 3^16

6^16 3^16