Logs after logs?

$\text{I could do all in mind... Believe me!}$

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\log_4\log_2\log_\sqrt{2}log_3 (x-2006) =0 \implies \log_2\log_\sqrt{2}log_3 (x-2006) =4^0=1 \implies \log_\sqrt{2}log_3 (x-2006) =2^1=2 $\implies log_3 (x-2006) =\sqrt2^2=2$ $\implies x-2006 =3^2=9\implies \boxed{x=2015} \, :)$

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$\text{The only property of logarithms used here is }\log_ab=c \implies b=a^c$

Finding the answer to this question only involves the knowledge of the two basic properties of logarithms. $log_{a}a = 1$ and $lob_{a}1 = 0$ . In the unlikely case of forgetting that, one can always picture the graph of the logarithmic function. And, one could always make things simpler by using $log_{a^{b}}c = \frac{1}{b}log_{a}c$ . You can compute the answer easily on your mind.

@Harshi Singh ...... sorry i found that question wrong to post the comments.....Sorry i need to again change it.....It was due to that oversmart Kshitiz........hell the man whom i hate most....so for one more time BEST OF LUCK FOR ASAT.