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Thread: Logs/Equations

  1. #1
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    Logs/Equations

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

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

    $\displaystyle 4^x-2^{x+1}=3$

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

    $\displaystyle 4^x-2^x+2-3=0$

    $\displaystyle 4^x-2^x -1= 0$

    $\displaystyle 2^x-1=0$

    $\displaystyle Let 2^x = y.$

    $\displaystyle y-1=0$
    $\displaystyle y= 1$

    $\displaystyle 2^x = 1$

    $\displaystyle 2^0=1$

    $\displaystyle x = 0$
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by A Beautiful Mind View Post
    The problem is: $\displaystyle 4^x-2^{x+1}=3$

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

    $\displaystyle 4^x-2^{x+1}=3$

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

    $\displaystyle 4^x-2^x+2-3=0$

    $\displaystyle 4^x-2^x -1= 0$

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

    $\displaystyle Let 2^x = y.$

    $\displaystyle y-1=0$
    $\displaystyle y= 1$

    $\displaystyle 2^x = 1$

    $\displaystyle 2^0=1$

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

    $\displaystyle -2^{x+1} = -(2^{x+1}) = -2^x-2$

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

    $\displaystyle 2^{2x}-2^x-5=0$

    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 $\displaystyle x\!:\;\;4^x-2^{x+1}\:=\:3$

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

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

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

    Therefore: .$\displaystyle 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
    MHF Contributor Matt Westwood's Avatar
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    Quote Originally Posted by A Beautiful Mind View Post
    The problem is: $\displaystyle 4^x-2^{x+1}=3$

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

    $\displaystyle 4^x-2^{x+1}=3$

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

    $\displaystyle 4^x-2^x+2-3=0$

    $\displaystyle 4^x-2^x -1= 0$

    $\displaystyle 2^x-1=0$

    $\displaystyle Let 2^x = y.$

    $\displaystyle y-1=0$
    $\displaystyle y= 1$

    $\displaystyle 2^x = 1$

    $\displaystyle 2^0=1$

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

    $\displaystyle - 2^{x+1}= -2^x+2^1$

    is actually wrong, should be:

    $\displaystyle - 2^{x+1} = - (2^x)(2^1)$
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