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I just found an inequality in my math book, how do I solve this?

$\log_2 3 > \log_3 4$

What I did:

$\log_2 3 > 2\log_3 2$

$\frac{1}{\log_3 2} > 2\log_3 2$

..and here is like $\frac{1}{a} > 2a$ and is impossible :(

How can I solve this?

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closed as off-topic by Andrés E. Caicedo, Xander Henderson, Don Thousand, user10354138, José Carlos Santos Oct 28 '18 at 23:53

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$\log_23$ vs $\log_34$

Multiply each term by $2$:

$2\log_23$ vs $2\log_34$

Apply logarithm rules:

$\log_23^2$ vs $\log_34^2$

Simplify:

$\log_29$ vs $\log_316$

Conclude:

$\log_29>\log_28=3=\log_327>\log_316$


Hence $\log_23>\log_34$

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  • $\begingroup$ @Macavity: Like it says in the answer - "logarithm rule" (coefficient outside the log can be substituted for an exponent inside the log). $\endgroup$ – barak manos Sep 27 '16 at 14:32
  • $\begingroup$ @Macavity: You're right, I'm wrong, will delete the answer... $\endgroup$ – barak manos Sep 27 '16 at 14:36
  • $\begingroup$ @Macavity: Damn, I thought you were the OP. Will flag it to someone who can unaccept, thanks... $\endgroup$ – barak manos Sep 27 '16 at 14:38
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    $\begingroup$ @MathematicianByMistak. What is nice in the present case is your user name ! Please, take into account that, if I was given a dime for each of my mistakes, I should be a millionaire ! Cheers :-) $\endgroup$ – Claude Leibovici Sep 27 '16 at 14:59
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    $\begingroup$ @barakmanos Nicely done +1! $\endgroup$ – Macavity Sep 27 '16 at 17:16
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Actually, $\frac1a > 2a$ is not impossible because then $a^2 < \frac12$ provided that $a > 0$. This means $\frac{-1}{\sqrt2} < a < \frac1{\sqrt2}$. This makes sense because $log_32 \approx 0.6309$, which is between $\frac{-1}{\sqrt2} \approx -0.7071$ and $\frac1{\sqrt2} \approx 0.7071$. I have no idea how to solve it because there is no variable to solve for, but rather it is a true statement:$$log_23 \approx 1.5850 > log_34 \approx 1.2619$$

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For any real numbers $x > 1$ and $y > 1$, and any positive integers $m$ and $n$, $$ \begin{Bmatrix} \log_x y > \frac{m}{n} \\ \log_x y = \frac{m}{n} \\ \log_x y < \frac{m}{n} \end{Bmatrix} \text{ according as } \begin{Bmatrix} x^m < y^n \\ x^m = y^n \\ x^m > y^n \end{Bmatrix}. $$ Taking $m = 3$ and $n = 2$, we have $2^3 = 8 < 9 = 3^2$, whereas $3^3 = 27 > 16 = 4^2$, and therefore $$ \log_2 3 > \frac{3}{2} > \log_3 4. $$

In this case - the value of $n$ being so small - one can easily check the result by numerical calculation: $$ 2^{3/2} = 2\sqrt{2} < 3, \text{ whereas } 3^{3/2} = 3\sqrt{3} > 4. $$

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