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Math Help - 2n=128

  1. #1
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    2n=128

    Hi! Great site. How would I solve for n if n is an exponent?

    2n=128
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rune2402 View Post
    Hi! Great site. How would I solve for n if n is an exponent?

    2n=128
    express 128 as 2 raised to some power, then equate the powers
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    Find the prime factorization of 128 and see how many twos there are in it.
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  4. #4
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    O yes I understand that , but

    I understand that, but how do I isolate n?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rune2402 View Post
    Hi! Great site. How would I solve for n if n is an exponent?

    2n=128
    Quote Originally Posted by SnipedYou View Post
    Find the prime factorization of 128 and see how many twos there are in it.
    Quote Originally Posted by rune2402 View Post
    I understand that, but how do I isolate n?
    2^n = 128

    Have you tried to find the prime factorization of 128, as SnipedYou suggested? This should answer your question. If not, post your factorization of 128 and I'll let you know how to proceed from there.

    -Dan
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  6. #6
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    Can anyone do the math so I can see

    Can someone write the math so i can see? I'd like to be able to do this for different problems.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rune2402 View Post
    I understand that, but how do I isolate n?
    i told you. when you are able to express 128 as a number with base 2 (you can use SnipedYou suggestion to help you in this endeavor, but it is a small enough number to do by trial and error pretty quickly) just equate the powers. this will isolate n in a new equation
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  8. #8
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    Quote Originally Posted by rune2402 View Post
    Hi! Great site. How would I solve for n if n is an exponent?

    2n=128
    Use logarithms

    2^n = 128
    Log(2^n) = Log(128)
    n*Log(2) = Log(128)
    n = log(128) / Log(2)
    n = 7 -------------------------answer.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ticbol View Post
    Use logarithms

    2^n = 128
    Log(2^n) = Log(128)
    n*Log(2) = Log(128)
    n = log(128) / Log(2)
    n = 7 -------------------------answer.
    this is an Elementary/Middle school poster, he/she probably does not know how to use logs yet.

    note that the prime factorization of 128 will show that:

    128 = 2*2*2*2*2*2*2 = 2^7

    so 2^n = 2^7

    since the bases are the same, the powers must be the same, thus we can equate them. and we get:

    n = 7
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  10. #10
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    thanks Ticbol

    Thank Ticbol. Thank you for the help.
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  11. #11
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    I told you...

    I told you...?

    Do you talk like that to your mom too or just people on the forum? Actually, I am in the 6th grade. I don't like rude people like that guy. So rude. I told you?
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  12. #12
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    Quote Originally Posted by rune2402 View Post
    I told you...?

    Do you talk like that to your mom too or just people on the forum? Actually, I am in the 6th grade. I don't like rude people like that guy. So rude. I told you?
    Some of the helpers/masters here sometimes think the askers are as good as they are. But some helpers too take the pains to explain to the most Math-challenged askers even if those in the first group might snicker at it.

    Helping those who seek help and solving the Problems only are two different things. It is more difficult to explain as detailed as possible. Solving is easy.

    Don't be disappointed in this your first try. Ask some more, anytime.
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  13. #13
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    Thanks Ticbol

    Thanks Ticbol. See ya later.
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rune2402 View Post
    I told you...?

    Do you talk like that to your mom too or just people on the forum? Actually, I am in the 6th grade. I don't like rude people like that guy. So rude. I told you?
    Quote Originally Posted by ticbol View Post
    Some of the helpers/masters here sometimes think the askers are as good as they are. But some helpers too take the pains to explain to the most Math-challenged askers even if those in the first group might snicker at it.

    Helping those who seek help and solving the Problems only are two different things. It is more difficult to explain as detailed as possible. Solving is easy.

    Don't be disappointed in this your first try. Ask some more, anytime.
    i see nothing rude about my response. i did not just say "i told you" i proceeded to explain further. furthermore, i gave a simpler solution than ticbol, again with explanation. the helpers here don't want to just give out solutions, they want the student to actually learn, so the often give hints and wait for the poster to try them. you showed no indication that you tried the hints that were given, which was the reason for my response. just asking for solutions won't help you. and as for thinking that other users are as "good as i am," i am guilty of no such thing, and frankly, i do not think any other helper here is, maybe other users, but no helper believes that.
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  15. #15
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    Hey Just kidding.

    I was just kidding. Hey Jamaica is very nice. Ya Mon.

    This is funny though:
    The Following 2 Users Say Thank You to Jhevon For This Useful Post: SnipedYou (Yesterday), topsquark (Yesterday)

    Heheh. If I say thank you do I get a cookie? Heheh.


    Thank you. Sorry for misunderstanding. I'm off to have a red stripe.
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