## Precalculus (6th Edition) Blitzer

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RECALL: (i) $\log_b{b}=1$ (ii) $\log_b{b^x}=x$ (iii) $\log_b{1} = 0.$ Note that $32=2^5$. Thus, the given expression can be written as $\log_5{(\log_2{2^5})}.$ Use rule (ii) above to obtain: $\log_5{(\log_2{2^5})}=\log_5{5}.$ Use rule (i) above to obtain $\log_5{5} = 1.$