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I'm trying to solve this problem:

If $\log_{27}(a)=b$, find $\log_{\sqrt[6]{a}}\sqrt{3}$

However, I'm unable to see any connection in those given information. How can I solve this logarithm?

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closed as off-topic by Henrik, user99914, Xander Henderson, Namaste, user223391 Jul 31 '18 at 1:08

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    $\begingroup$ I'm unable to see any connection in those given information What do you know about logarithms change of base? $\endgroup$ – dxiv Jul 29 '18 at 6:58
  • $\begingroup$ @dxiv you mean this notation: $log_ab=\frac{1}{log_ba}$? $\endgroup$ – Steve Jul 29 '18 at 7:03
  • $\begingroup$ @Steve: That's not notation, that's a formula - that seems relevant in this case - but I would just use $\log_a b=\frac{\log b}{\log a}$ (from which your formula follows). $\endgroup$ – Henrik Jul 29 '18 at 7:07
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    $\begingroup$ @Steve And more generally ${\log _{a}b={\frac {\log _{c}b}{\log _{c}a}}.\,}$ $\endgroup$ – dxiv Jul 29 '18 at 7:09
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    $\begingroup$ Why don't you just solve for $a$ and substitute? $\endgroup$ – Miksu Jul 29 '18 at 7:58
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Given $\log_{27}(a) = b$ then $a = 27^b$. Then: $$\sqrt[6]{a} = (27^b)^\frac{1}{6} = (3^{3 \cdot \frac{1}{6}})^b = 3^\frac{b}{2}$$ $$\log_{\sqrt[6]{a}}x = \log_{3^\frac{b}{2}}x = \frac{\log_{3}x}{\log_{3}3^\frac{b}{2}} = \frac{2}{b}\log_{3}x$$ Since $x = \sqrt{3}$ then $$\frac{2}{b}\log_{3}x = \frac{2}{b}\cdot\frac{1}{2} = \frac{1}{b}$$

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