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log base 6 [log base 5 (log base 3 x)] = 0
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\[\log_{6} [\log_{5} (\log_{3}x) ]=0 \]
pealing off one at a time, this tells you \[\log_5(\log_3(x))=1\] as a first step
this in turn means \(\log_3(x)=5\)
and finally this gives \(x=3^5\)
Wait, how did you know that \[\log_{5} (\log_{2} (x))=1\] ?
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And what happened to\[\log_{6}\]
\[\log_{a}x=b \] means \[a ^{b}=x\]
The great one used that principle 3 times.
The great one...haha. :P
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