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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 :
    $\displaystyle - \frac 1x \ln 8=\ln \frac{25}{2}=\ln 25 - \ln 2$

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

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

    The equation is now :

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

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

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

    $\displaystyle x=\frac{1}{\frac 13-2\frac{\ln 5}{3 \ln 2}} \approx -0.82330362225011$
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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?
    $\displaystyle 8^{\frac{-1}{x}}=\frac{25}{2}$

    Take everything to the "x" power.

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

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

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

    Take the natural log of both sides

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

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

    Another approach . . .


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

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

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

    Hence: .$\displaystyle \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.


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


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

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


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

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


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

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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 $\displaystyle 10^{1 - 1} + 1 = 10^0 + 1 = 2$.

    So, we really have

    $\displaystyle
    \begin{array}{ccc}
    10^{x-1} + 1 &=& 2\\
    2^{2x-1} &=& 2
    \end{array}\quad\Rightarrow\quad\boxed{x \:=\:1}
    $

    Of course, you still had the right solution.

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