Solve:
\[10^{\log_{p}(\log_{q}(\log_{r}x)} = 1\]and \[\log_{q}(\log_{r}(\log_{p}x))=0\] Then p equals to - a)r^{q/r} c)1 b)rq d)r^{r/q}
@Callisto @ganeshie8
its A
\[\log_b1=0\]
@nitz its A .How?
\[ 10^{\log_p (\log_q (\log_r x) ) } = 1\] \[ \log_p (\log_q (\log_r x) ) = 0\] \[ p^0 = \log_q (\log_r x)\] \[ \log_q (\log_r x) = 1\] \[q = \log_r x\] \[r^q = x\] then \[\log_q (\log_r (\log_p x) ) = 0\] \[q^0 = \log_r (\log_p x)\] \[\log_r (\log_p x) = 1\] \[r = \log_ p x\] \[p^r = x\] since both equations are equal to x i can equate them to one another \[p^r = r^q\] so i want p so i raise both sides to 1/r \[p = (r^q)^{1/r}\] \[p = r^{q/r}\] got it?
so yes it's A nitz was right
just tell me if something wasnt clear for you ^_^
got it.
great stuff @lgbasallote \[\log_bb=1\]
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