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Jack
Lv 4
Jack asked in Education & ReferenceHomework Help · 1 decade ago

Algebra Advanced Exponents, tricky, first right answer 10pts?

Asked this before but had a bad link, sorry

I'm just testing yahoo answers to see who here actually knows what they're talking about (at least in higher algebra)

Here's a link to a moderately hard advanced algebra problem, it can be tricky if you think you know what your talking about when you don't. It basically focuses on people's understanding of exponents.

http://img146.imageshack.us/img146/9964/guessgw9.p...

I'll post an answer soon enough, Go for it!

I just don't like people posting answers to other questions when they don't understand the concepts

I'm getting an answer from this question even if its not the mathematical answer

Update:

good idea sushi, but can't combine it like that, they're bound by their exponents

Update 2:

I'll show the answer and explanation tomorrow morning if anyone's interested :P

Update 3:

No, no logs needed

Update 4:

Ohh Baby!!!

Nina wins :D

Answer:

http://img388.imageshack.us/img388/930/answeron4.p...

at least some people know their math

9 Answers

Relevance
  • 1 decade ago
    Favorite Answer

    2 x (2 ^2k + 2 ^2k) ^2

    = 2 x ( (4^k) ^2 + 2 (4^k)(4^k) + (4^k)^2)

    = 2 x (4^2k + 2(4^2k) + 4^2k)

    = 2 x 4^2k (1 +2+1+)

    = 2 x 4^2k (4)

    = 2x 2^4k x 2^2

    = 2^1+4k+2

    = 2^4k+3

    I wish I did it right.:)P

  • 1 decade ago

    Easy use logs, Log base 4k+4K, 4 then that equals log base 4k+4k to the x over 2 is equal to two. You could use base change to simplify it further but I dunno how simplified you want it. Dude you asked to simplify it and I used my knowledge in calculus. This is the calculus method of simplifying exponents. Logs are exponents or at least have similar traits.

    Source(s): Finite mathematics student
  • 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.

  • 1 decade ago

    8(16^k)

    Verify: If k = 5

    2(4^5 + 4^5)^2 = 8(16^5)

    8,388,608 = 8,388608

  • How do you think about the answers? You can sign in to vote the answer.
  • 1 decade ago

    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…..

  • Alian
    Lv 4
    1 decade ago

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

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

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

    = 2^3 * (((2^2)^k)^2)

    = 2^3 * 2^(4k)

    = 2^(3+4k)

    EDIT: Hehe. I posted too late. But it was right!

    Source(s): This had better be right... I go to an engineering college. ::wince::
  • 1 decade ago

    16^2k

  • 1 decade ago

    Ahh..I have a feeling it's probably wrong, but here's my answer:

    128^2k

    Ehh.. I dunno; first time answering a question in the homework help section (:

  • 1 decade ago

    GO DO YOUR OWN HOMEWORK, MY FRIEND. STOP BEING LAZY AND WISHING YOU COULD GET 100 PERCENT!

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