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    2^x+4^x+1=8

    How do you solve this?
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    (Original post by judogainz)
    2^x+4^x+1=8

    How do you solve this?
    Having everything with a base of 2^... may help.
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    (Original post by SeanFM)
    Having everything with a base of 2^... may help.
    taht was what is was thinking how would i get the other one to 2?
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    (Original post by judogainz)
    taht was what is was thinking how would i get the other one to 2?
    Law of indices.

    How can 4^x be written in terms of 2^...? (Hint: 4=2^2)
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    (Original post by SeanFM)
    Law of indices.

    How can 4^x be written in terms of 2^...? (Hint: 4=2^2)
    i know that you square it, but would it go to 2^x+3^2(x+1) ?
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    (Original post by judogainz)
    i know that you square it, but would it go to 2^x+3^2(x+1) ?
    The 3 should be a 2, but yes, well done :borat:
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    (Original post by SeanFM)
    The 3 should be a 2, but yes, well done :borat:
    so you ignore the base and just use the pwoers

    Im left with x+2(x+1)=8

    do i just solve it from here?
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    (Original post by judogainz)
    so you ignore the base and just use the pwoers

    Im left with x+2(x+1)=8

    do i just solve it from here?
    You can't quite jump like that

    So the things with x in are now of the form 2^...

    Giving you 2^x + 2^(x+2) - 8 = 0, and you know that 2^(x+2) = 2^x * 2^2 by law of indices. After you've changed that, can you think of how to solve it? (Hint: a substitution turns it into a quadratic).
 
 
 
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Updated: November 12, 2015
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