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Thread: SOLVE FOR x please....

  1. #1
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    SOLVE FOR x please....

    e4x+1= 2ex
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  2. #2
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    Re: SOLVE FOR x please....

    Quote Originally Posted by Scoplex View Post
    e4x+1= 2ex
    Can you solve $\displaystyle 3x+1=\log(2)~?$
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  3. #3
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    Re: SOLVE FOR x please....

    x = [log2 - 1]/3 ???

    Does this mean that log(2) is equal to 2e?
    I would then do the following:
    4x+1=x
    3x=-1
    x=-1/3 ??? Is this then correct?

    This logs and stuff get my head cooking sometimes.
    Thanks in advance.
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  4. #4
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    Re: SOLVE FOR x please....

    Quote Originally Posted by Scoplex View Post
    x = [log2 - 1]/3 ???CORRECT

    Does this mean that log(2) is equal to 2e?
    What do you mean by that?

    The $\displaystyle \log(2)$ is just $\displaystyle \log(2)$, no more, no less.
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  5. #5
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    Re: SOLVE FOR x please....

    You already have the answer as x = [log(2) - 1]/3.
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  6. #6
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    Re: SOLVE FOR x please....

    Hello, Scoplex!

    $\displaystyle \text{Solve for }x\!:\;\;e^{4x+1} \:=\: 2e^x$

    Take logs: .$\displaystyle \ln\left(e^{4x+1}\right) \:=\:\ln(2e^x)$

    . . . . . . $\displaystyle (4x+1)\ln(e) \:=\:\ln(2) + \ln\left(e^x\right)$

    . . . . . . $\displaystyle (4x+1)\ln(e) \:=\:\ln(2) + x\ln(e)$ . . . Note that $\displaystyle \ln(e) = 1$

    n . . . . . . . . . $\displaystyle 4x + 1 \:=\:\ln(2) + x$

    n . . . . . . . . . $\displaystyle 3x + 1 \:=\:\ln(2)$

    . . . . . . . . . . . . . $\displaystyle 3x \:=\:\ln(2) - 1$

    . . . . . . . . . . . . . .$\displaystyle x \:=\:\dfrac{\ln(2) - 1}{3}$
    Thanks from sbhatnagar
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