## Precalculus (6th Edition) Blitzer

$0$
RECALL: (i) $\log_b{b}=1.$ (ii) $\log_b{b^x}=x.$ (iii) $\log_b{1} = 0.$ (iv) $\ln{x} = \log_e{x}.$ (v) $\log{x} = \log_{10}{x}.$ Use rule (iv) above to obtain: $\log{(\ln{e})}=\log{(\log_e{e})}.$ Use rule (i) above to obtain $\log{(\log_e{e})}=\log{1}.$ Use rule (v) to obtain $\log{1}=\log_{10}{1}.$ Use rule (iii) above to obtain $\log_{10}{1} = 0.$ Therefore $\log{(\ln{e})}=0$.