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Show that

$$1-\frac{2}{3}^{\frac{1}{\log_2{\frac{2}{3}}}}=\frac{1}{2}.$$

I was trying to calculate it but I failed anytime. I would be grateful for help.

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  • $\begingroup$ $a^{1/\log_2a}=2^{\log_2a/\log_2a}=2$, so the statement is wrong. $\endgroup$
    – user65203
    Dec 18, 2019 at 15:43

2 Answers 2

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Use the following formulas:

$$\log_a b = \frac{1}{\log_b a}$$ and $$a^{\log_a b} = b$$.

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  • $\begingroup$ Ok, now I get it! Thanks. $\endgroup$
    – Uhans
    Dec 18, 2019 at 15:23
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Since $x^{\frac{1}{\log_2 x}} = x^{\log_x 2} = 2$ then for $x = \frac{2}{3}$ we get $$(\frac{2}{3})^{\frac{1}{\log_2 \frac{2}{3}}} = 2$$

So $$1 - (\frac{2}{3})^{\frac{1}{\log_2 \frac{2}{3}}} = -1$$

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