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Math Help - Need help with exponential equations

  1. #1
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    Question Need help with exponential equations

    2x + 2x+1 + 2x+2 = 56

    The solution is the following:

    2x + 2 * 2x+ 4 * 2x = 56

    Why is that? I don't understand how it came to this. Where did the 4 come from? How come there's so many 2s?

    7 * 2x = 56

    2x = 8

    How did 56 turn into 8? Shouldn't it be 82 instead of just 8?

    2x = 23

    x = 3

    I only understand the last two steps. Please help, I need to know this (and more) for a test. This isn't the only equation that confuses me, but it would help to know what lead to the final solution.
    Please reply,
    thanks.
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  2. #2
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    Re: Need help with exponential equations

    We are given:

    2^x+2^{x+1}+2^{x+2}=56

    Using the property of exponents a^b\cdot a^c=a^{b+c} we may rewrite the equation as:

    2^x+2^1\cdot2^x+2^2\cdot2^x=56

    2^x+2\cdot2^x+4\cdot2^x=56

    2^x(1+2+4)=56

    7\cdot2^x=56

    Divide through by 7:

    2^x=8=2^3

    Hence:

    x=3
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    Re: Need help with exponential equations

    Thanks, now I know how the conversion to multiplying occured.

    However, I don't understand this part:

    2^x(1+2+4)=56

    Wouldn't

    2^x+2\cdot2^x+4\cdot2^x=56

    be 2x + 4x + 8x = 56?

    How did you come up with the "1"?

    Also, about dividing through by 7: 7 is not their common denominator - 2 is not in seven (at least not without a decimal). How does it stay unchanged? Is it because of the X?

    Thanks
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Need help with exponential equations

    Quote Originally Posted by dirtyharry View Post
    ...

    I don't understand this part:

    2^x(1+2+4)=56

    Wouldn't

    2^x+2\cdot2^x+4\cdot2^x=56

    be 2x + 4x + 8x = 56?

    How did you come up with the "1"?
    Note that:

    2^x+2\cdot2^x+4\cdot2^x=56

    is not the same as:

    2^x+(2\cdot2)^x+(4\cdot2)^x=56

    To make things simpler to see, let u=2^x then we have:

    u+2u+4u=56

    Now combine like terms on the left:

    7u=56

    Divide both sides by 7:

    u=8

    Back-substitute for u:

    2^x=8=2^3

    You see, when I factored in my first post, I used the fact that each term on the left has 2^x as a factor.
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  5. #5
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    Re: Need help with exponential equations

    Quote Originally Posted by MarkFL2 View Post
    Note that:

    2^x+2\cdot2^x+4\cdot2^x=56

    is not the same as:

    2^x+(2\cdot2)^x+(4\cdot2)^x=56

    To make things simpler to see, let u=2^x then we have:

    u+2u+4u=56

    Now combine like terms on the left:

    7u=56

    Divide both sides by 7:

    u=8

    Back-substitute for u:

    2^x=8=2^3

    You see, when I factored in my first post, I used the fact that each term on the left has 2^x as a factor.
    Thanks
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