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  1. #1
    Newbie yuriythebest's Avatar
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    new question

    sorry that I couldn't make the title more descriptive.

    anyhow I've almost solved an equation and I now have



    what's next? how do I get x out of this?


    also

    if I have 10^x - 5*4^x + 10=0

    what can I possibly do next?
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  2. #2
    Moo
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    Hello

    Quote Originally Posted by yuriythebest View Post
    sorry that I couldn't make the title more descriptive.

    anyhow I've almost solved an equation and I now have



    what's next? how do I get x out of this?
    Taking the logarithms :
    - \frac 1x \ln 8=\ln \frac{25}{2}=\ln 25 - \ln 2

    But 8=2^3 \implies \ln 8=3 \ln 2 (it's more beautiful )

    And 25=5^2 \implies \ln 25=2 \ln 5

    The equation is now :

    -\frac 1x (3\ln 2)=2 \ln 5-\ln 2

    - \frac 1x=2 \frac{\ln 5}{3\ln 2}-\frac 13

    \frac 1x=\frac 13-2\frac{\ln 5}{3 \ln 2}

    x=\frac{1}{\frac 13-2\frac{\ln 5}{3 \ln 2}} \approx -0.82330362225011
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  3. #3
    GAMMA Mathematics
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    Quote Originally Posted by yuriythebest View Post
    sorry that I couldn't make the title more descriptive.

    anyhow I've almost solved an equation and I now have



    what's next? how do I get x out of this?


    also

    if I have 10^x - 5*4^x + 10=0

    what can I possibly do next?
    8^{\frac{-1}{x}}=\frac{25}{2}

    Take everything to the "x" power.

    8^{\frac{-x}{x}}=\left(\frac{25}{2}\right)^x

    8^{-1}=\left(\frac{25}{2}\right)^x

    \frac{1}{8}=\left(\frac{25}{2}\right)^x

    Take the natural log of both sides

    \ln \left(\frac{1}{8}\right)=x \ln \left(\frac{25}{2}\right)

    Can you solve that?
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  4. #4
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    Hello, yuriythebest!

    Another approach . . .


    8^{-\frac{1}{x}} \:=\:\frac{25}{2}

    Take reciprocals: . 8^{\frac{1}{x}} \:=\:\frac{2}{25} \:=\:\frac{8}{100}

    Take logs (base 10): . \log\left(8^{\frac{1}{x}}\right) \:=\:\log\left(\frac{8}{100}\right)\quad\Rightarro  w\quad \frac{1}{x}\log(8) \:=\:\log(8) - 2

    Hence: . \frac{1}{x} \:=\:\frac{\log(8) - 2}{\log(8)} \quad\Rightarrow\quad\boxed{ x \;=\;\frac{\log(8)}{\log(8) - 2}}

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  5. #5
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    Heloo again, yuriythebest!

    The second problem is rather bizarre.
    I will assume we want integer solutions.


    10^x - 5*4^x + 10\:=\:0
    I found no direct approach to it . . .


    We have: . 10^x + 10 \:=\:5\cdot4^x \quad\Rightarrow\quad 10\left(10^{x-1}+1\right) \:=\:5\cdot2^{2x}

    Divide by 10: . 10^{x-1}+1 \;=\;2^{2x-1}


    The left side is of the form: . \hdots\:1.001,\:1.01,\:1.1,\:{\color{red}1},\:11,\  :101,\:1001,\:\hdots

    The right side is of the form: . \hdots\:\frac{1}{8},\:\frac{1}{4},\:\frac{1}{2},\:  {\color{red}1},\:2,\:4,\:8,\:\hdots


    They are equal only when: . \begin{array}{ccc}10^{x-1} + 1 &=& 1 \\ 2^{2x-1} &=& 1\end{array}\quad\Rightarrow\quad\boxed{x \:=\:1}

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  6. #6
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    As usual, Soroban, your post is very useful. However, the left side of your equation is never actually 1. The highlighted element should be 10^{1 - 1} + 1 = 10^0 + 1 = 2.

    So, we really have

    <br />
\begin{array}{ccc}<br />
10^{x-1} + 1 &=& 2\\<br />
2^{2x-1} &=& 2<br />
\end{array}\quad\Rightarrow\quad\boxed{x \:=\:1}<br />

    Of course, you still had the right solution.

    Incidentally, the other solution should be about x\approx1.364841930549761, determined by Newton's method.
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