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Can I further simplify this logarithmic expression somehow in this case, or is it already the point when I take a calculator?

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Blue
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brilliant
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    Well, you can rewrite it in terms of $\log 2$ and $\log 3$ if you prefer, but I'm not sure that's any simpler. – lulu Jun 15 '22 at 13:46
  • @lulu - "... you can rewrite it in terms of log2 and log3 if you prefer" - How could I do that? – brilliant Jun 15 '22 at 13:50
  • $\log 6 = \log 2+\log 3$ and $\log 4=2\log 2$. Definitely worth trying but, as I say, I don't think the resulting expression is any simpler. – lulu Jun 15 '22 at 13:58
  • @lulu - Ah! I got it! Thank you. – brilliant Jun 15 '22 at 14:00
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    $\log 6-\log 4=\log{6\over 4}=\log{3\over 2}$, ${\log 6 \over \log {3\over 2}}=\log_{3\over 2} 6$, $-2\log_a 6=\log_a 6^{-2}=\log_a {1\over 36}$, then answer is $\log_{3\over 2} {1\over 36}$, which is the same as $\log_{2\over 3} 36$ – Ivan Kaznacheyeu Jun 15 '22 at 15:28

2 Answers2

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Note that $\log 6=\log (2\cdot 3) =\log2+\log3$, and $\log 4=2\log 2$. So x could be simplified to $\frac{-2\log 6}{\log \frac{3}{2}}$

Guillermo García Sáez
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If you prefer an expression with ust one log-term in it, divide by $\log 6$:

$$\begin{align} x &= \frac{-2\log 6}{\log6 - \log4} \\ &= \frac{-2}{1-\log_6 4} \\ &= \frac{2}{\log_6 4 -1} \\ \end{align}$$

If you prefer log in the numerator, you can absorb $-1=\log_6 6$ into the other log term and use $\log_a b = 1/\log_b a$:

$$\begin{align} x &= \frac{2}{\log_6 4 -\log_6 6} \\ &= \frac{2}{\log_6 (4/6)} \\ &= 2 \log_{2/3} 6 = -2\log_{1.5} 6\\ \end{align}$$

emacs drives me nuts
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