My answer: Fuck this just give me a rifle and point me towards the enemy lines!

  • Achyu@lemmy.sdf.org
    ·
    7 months ago

    As the other comrade said, you've got the answers.

    For (2)
    - 2t2 + 7t -3 > 0
    Or 2t2 - 7t -3 <0
    We can find the roots and get 2t2 - 7t -3 as (t - 3) (2t - 1) < 0

    From this we can see that it's less than zero when t < 3, because the (t - 3) term becomes negative But if t < 1/2, both terms become negative and the equality is not valid.

    Thank you. This shows both bounds. I had overlooked that and cancelled the (2t - 1) in a previous step.

    So the answer would be 1/2 < ln x < 3
    Thanks again

    For (3) we have log3 x30 = 30 ln x / ln 3
    Using the property loga X = logb X / logb a

    A question: How did you type of out the equations? LaTeX?
    Is there any simple/foss app for it on Android?

    • AernaLingus [any]
      ·
      7 months ago

      Thank you! They both seem so simple now that I've seen your explanation, but that's always how it is with these things; I'm sure we've all had the experience of nodding along to something in class only to be completely flummoxed by the same material when doing the homework later that day.

      With (2), I had a feeling that I needed to factor, but I've never been that great at factoring quadratics when the second degree's coefficient isn't 1, and even if I had succeeded in factoring I don't think I would have remembered the logic to work through to get the final inequality. But now I understand thanks to your explanation!

      solution for (2)

      Show

      With (3), for some silly reason I thought that the rule with logaxb = b ⋅ loga x only applied when x = a (no idea why—obviously in that case you just have logaab = b) ...as I said, it's been a while since I did this stuff regularly. That was the key I needed to finish that one!

      solution for (3)

      Show

      Yup, I used LaTeX! I'm not personally familiar with an app targeted at Android, but I used the web-based Overleaf, which is FOSS (AGPLv3). While the website I linked to does require the creation of a free account (which was all I ever needed for my simple undergraduate homework purposes--I think the premium features are only relevant for large projects/teams), you can even self-host it if you're into that sort of thing.

      Incidentally, here's the LaTeX markup for the problems I worked (nothing fancy, just using the basic math packages with the main feature being the \align environment which allows me to center the equations around the = or >/< symbols).

      LaTeX markup 
      \documentclass{article}

\usepackage{amsmath, amsthm, amssymb, amsfonts, mathtools}

% ------------------------------------------------------------------------------

\begin{document}
% (2)
Substitute \(t = \log_{e}{x} = \ln{x}\).
\begin{align*}
(0.2^{(6t - 3)})^{1/t} &> (0.008^{2t - 1})^{1/3} \\ 
0.2^{(6t - 3)/t} &> (0.2^{3})^{(2t - 1)/3} \\ 
0.2^{(6t - 3)/t} &> 0.2^{3 \cdot [(2t - 1)/3]} \\ 
\frac{6t - 3}{t} &> 3 \cdot \frac{2t - 1}{3} \\ 
t \cdot \frac{6t - 3}{t} &> (2t - 1) \cdot t \\ 
6t - 3 &> 2t^2 - t \\ 
-2t^{2} + 7t - 3 &> 0 \\
2t^2 - 7t + 3 &< 0 \\
(2t - 1)(t - 3) &< 0 \\
\end{align*}

When \(t < \frac{1}{2}\) or \(t > 3\), the product of the factored terms is positive, so the inequality doesn't hold.\\
When \(\frac{1}{2} < t < 3\), the product of the factored terms is negative, so the inequality holds.\\
Substituting the original value of \(t\), we have that

\begin{equation*}
    \boxed{\frac{1}{2} < \ln{x} < 3}
\end{equation*}

%(3)
\begin{align*}
    3^{\ln{x} + \ln{x^0} + \ln{x^0} + \cdots + \ln{x^0}} &= 27x^{30} \\
    3^{\ln{x} + \ln{1} + \ln{1} + \cdots + \ln{1}} &= 27x^{30} \\
    3^{\ln{x} + 0 + 0 + \cdots + 0} &= 27x^{30} \\
    3^{\ln{x}} &= 27x^{30} \\
    \log_3{3^{\ln{x}}} &= \log_3{27x^{30}} \\
    \ln{x} &= \log_3{3^{3}} + \log_3{x^{30}} \\
    \ln{x} &= 3 + \log_3{x^{30}} \\
    \ln{x} &= 3 + \log_3{x^{30}} \\
    \ln{x} &= 3 + \frac{\ln{x^{30}}}{\ln{3}} \\
    \ln{x} &= 3 + \frac{30\ln{x}}{\ln{3}} \\
    \ln{3} \cdot \ln{x} &= 3\ln{3} + 30\ln{x} \\
    0 &= 3\ln{3} + 30\ln{x} - \ln{3} \cdot \ln{x} \\
    (30 - \ln{3})\ln{x} &= -3\ln{3} \\
    \ln{x} &= \frac{-3\ln{3}}{30 - \ln{3}}\\
    e^{\ln{x}} &= e^{\frac{-3\ln{3}}{30 - \ln{3}}}\\
    \Aboxed{x &= e^{\frac{-3\ln{3}}{30 - \ln{3}}}}
\end{align*}

\end{document}
      • Achyu@lemmy.sdf.org
        ·
        7 months ago

        Thank you. Since I have MS word, I sort-of make-do with it's equation editor,

        Been interested in LaTeX for a while, so thanks for this.

        • AernaLingus [any]
          ·
          7 months ago

          No problem! I think it's definitely worth learning the basics, which I don't think are too bad, and are really plenty unless you have very specific needs or want to make complex diagrams. And regardless, the wizards on StackOverflow are always there to help.

          By the way, I just noticed that there's an issue with the markup I posted due to a known Lemmy bug where ^ gets replaced with the sup HTML tag even in code blocks. I'll post a proper Pastebin or something when I get the chance!

            • AernaLingus [any]
              ·
              7 months ago

              No problem! Yeah, as you can see it's a bit different--basically, you've got commands that start with backslashes, which can have one or more arguments contained in curly brackets. The most important thing to understand is probably math mode which is what enables all the fancy mathematical typesetting.

              Here's the Pastebin with the correct markup for my earlier comment:

              https://pastebin.com/q3Sww7QM