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Math Help - Logs/Equations

  1. #1
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    Logs/Equations

    The problem is: 4^x-2^{x+1}=3

    Here's the work I've tried doing on it:

    4^x-2^{x+1}=3

    I split the 2^{x+1} up into -2^x+2^1.
    Then I...

    4^x-2^x+2-3=0

    4^x-2^x -1= 0

    2^x-1=0

    Let 2^x = y.

    y-1=0
    y= 1

    2^x = 1

    2^0=1

    x = 0
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  2. #2
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    Quote Originally Posted by A Beautiful Mind View Post
    The problem is: 4^x-2^{x+1}=3

    Here's the work I've tried doing on it:

    4^x-2^{x+1}=3

    I split the 2^{x+1} up into -2^x+2^1.
    Then I...

    4^x-2^x+2-3=0

    4^x-2^x -1= 0

    2^x-1=0 me: (Different bases mean you can't take it away like that )

    Let 2^x = y.

    y-1=0
    y= 1

    2^x = 1

    2^0=1

    x = 0
    4^x-2^{x+1}=3

    -2^{x+1} = -(2^{x+1}) = -2^x-2

    The key to this question is recognising that 4 can be written as 2^2

    2^{2x}-2^x-5=0

    This is a quadratic in 2^x
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  3. #3
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    Hello, A Beautiful Mind!

    Your "split" is incorrect.


    Solve for x\!:\;\;4^x-2^{x+1}\:=\:3

    We have: . (2^2)^x - 2\!\cdot\!2^x - 3 \:=\:0 \quad\Rightarrow\quad 2^{2x} - 2\!\cdot\!2^x - 3 \:=\:0

    Factor: . \left(2^x + 1\right)\left(2^x - 3\right) \:=\:0

    Then: . \begin{array}{ccccccc}2^x+1 \:=\:0 & \Rightarrow & 2^x \:=\:1 & &\text{no real roots} \\<br />
2^x - 3 \:=\:0 & \Rightarrow & 2^x \:=\:3 & \Rightarrow & x \:=\:\log_2(3)<br />
\end{array}

    Therefore: . x \:=\:\log_2(3)\:\text{ or }\:\frac{\ln(3)}{\ln(2)}

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  4. #4
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    EDIT: Just saw Soroban's post.

    I was trying to use my teacher's method...lol.
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  5. #5
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by A Beautiful Mind View Post
    The problem is: 4^x-2^{x+1}=3

    Here's the work I've tried doing on it:

    4^x-2^{x+1}=3

    I split the 2^{x+1} up into -2^x+2^1.
    Then I...

    4^x-2^x+2-3=0

    4^x-2^x -1= 0

    2^x-1=0

    Let 2^x = y.

    y-1=0
    y= 1

    2^x = 1

    2^0=1

    x = 0
    Just in case - I don't think it was pointed out that:

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

    is actually wrong, should be:

    - 2^{x+1} = - (2^x)(2^1)
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