40points for right answer! Algebraic Tricky Problem?

2^(3+4k)

im not 100 per cant positive, but i think it is something like: 8 to the power of 2k + 8 to the power of 2k. which would just simplify to give 16 to the power of 2k. sorry if this is wrong :)

When you said advanced algebra, I was hoping I would get to prove a homomorphism or something fun.

OK, I'll nibble...

2^(4k+3)

70 miles / 1 hour = 200 miles / x hours 70 = 200 / x ──► x = 200 / 70

2(4^k + 4^k)^2

= 2*[2(4^k)]^2

= 2[4(4^2k)]

= 2[4^(2k + 1)]

= 2[2^(4k + 2)]

= 2^(4k + 3)

I didn't even look at the answer.

Question::: : 2(4^K + 4^K)^2 ------------- (1)

Solution :::: :

= 2( 2*4K)^2

= 2( 2^2*4^2K)

= 2^3 * 4^2K

= 2^3 * (2^2)^2K

= 2^3 * 2^4K

= 2^(3+4K)---------------------(2) ANSWER

Proof ::::

In eqn 1 Substitute K=2------------ 2048

In eqn 2 Substitue K=2------------ 2048

Tats it…..

im not sure this is right :]

64(exponent k to the 2 power)

16^2k

2^4k+3

look in your other question for how i got it

8(16^2k)