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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?

Prajwal Kansakar
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Steve
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    I'm unable to see any connection in those given information What do you know about logarithms change of base? – dxiv Jul 29 '18 at 06:58
  • @dxiv you mean this notation: $log_ab=\frac{1}{log_ba}$? – Steve Jul 29 '18 at 07:03
  • @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). – Henrik supports the community Jul 29 '18 at 07:07
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    @Steve And more generally ${\log _{a}b={\frac {\log _{c}b}{\log _{c}a}}.,}$ – dxiv Jul 29 '18 at 07:09
  • ok, I used the formula as follows: $\frac{log3}{log\sqrt[6]{a}}$, since $log\sqrt[6]{a}=\frac{loga}{6}$, I have this left: $\frac{6log3}{loga}$. What now? – Steve Jul 29 '18 at 07:39
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    Why don't you just solve for $a$ and substitute? – Miksu Jul 29 '18 at 07:58

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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 $$\frac{2}{b}\log_{3}x = \frac{2}{b}\cdot\frac{1}{2} = \frac{1}{b}$$

Giulio Scattolin
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