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If $\log_{30}3=a,\log_{30}5=b$ then show what $\log_{30}8$ is.

I am having trouble trying to get 8 to be some sequence with 5,30,3. Any hints?

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    $\begingroup$ Hint: $\log_{30} 2 = \log_{30} 30/15 = 1-(a+b)$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 25 '18 at 10:50
  • $\begingroup$ @GNUSupporter Yes, that is what I am trying to do! I don't see anything that can give 8 though. $\endgroup$ – RiktasMath Apr 25 '18 at 10:51
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    $\begingroup$ $\log x^a = a \log x$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 25 '18 at 10:55
  • $\begingroup$ @GNUSupporter I am also aware of that. $\endgroup$ – RiktasMath Apr 25 '18 at 10:56
  • $\begingroup$ @GNUSupporter Omg I see it $\endgroup$ – RiktasMath Apr 25 '18 at 10:56
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$ \log_{30}8=\log_{30}2^3=3\log_{30}2=3\log_{30}\frac{30}{5\ast3}$

$3\log_{30}30-(3\log_{30}5+3\log_{30}3)=3-3b-3a=3(1-b-a)$

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