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Solve $(\log_5 x)^2 - 3\log_5(x) + 2 = 0$.

I tried to solve this question but got stuck:

\begin{align} (\log_5 x)^2 - 3\log_5(x) + 2 &= 0 \\ (\log_5 x)^2 &= \log_5(x^3) - 2 \\ \log_{(\log_5 x)}(\log_5 x)^2 &= \log_{( \log_5 x)}(\log_5(x^3) - \log_5 25) \\ 2 &= \log_{(\log_5 x)}\left( \log_5\frac{x^3}{25} \right) \end{align}

Was my approach wrong? Where do I go from here?

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    $\begingroup$ $\log_5x=t$. This is a quadratic equation. $$(t-1)(t-2)=0$$ $\endgroup$ Jul 2, 2021 at 17:13
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    $\begingroup$ Let $y = \log_5(x).$ Then construct a quadratic equation in $(y)$. Then solve for $(y)$. Then convert your solution into a solution for $(x)$. $\endgroup$ Jul 2, 2021 at 17:15
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    $\begingroup$ Your calculations are correct but you are spinning your wheels without an end plan. $\log_{\log_a} \log_a K$ is probably a bad idea. I would suggest taking $5$ to each side of the equation (if solving as quadratic never occurs to us) but that leads us to $x^{\log_5 x} = \frac {x^3}{25}=\frac {x^3}{x^{\log_x 25}} = x^{3-\log_x 25}$ and $\log_5 x= 3 - \log_x 25= 3-2\log_x 5= 3 -\frac 2{\log_5 x}$ which will be just spinning wheels until it occurs to us to replace $u =\log_5 x$ and solving. Eventually it will need to hit us this is just a quadratic. $\endgroup$
    – fleablood
    Jul 2, 2021 at 17:43
  • $\begingroup$ As a side note, it is generally difficult to deal with $(\log x)^y$. $\endgroup$
    – Trebor
    Jul 2, 2021 at 17:43
  • $\begingroup$ Why @Trebor ? It's $\log x^{\log x^{\dots}}$ with height $y$ $\endgroup$ Jul 3, 2021 at 15:13

1 Answer 1

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This is just a quadratic equation, but instead of variables you have logarithms. You can let $u=\log_5x$ and get

$$u^2-3u+2=0$$

$$(u-2)(u-1)=0$$

from which $\log_5x=1$ and $\log_5x=2$.

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