$2^{1000}?$

$As\quad { 2 }^{ 1000 }\quad is\quad not\quad a multiple\quad of\quad 10,\quad it\quad follows\quad that,\\ { 10 }^{ m }<{ 2 }^{ 1000 }<{ 10 }^{ m+1 },where{ 10 }^{ m }\quad contains\quad m+1\quad digits.\\ Solving\quad { 2 }^{ 1000 }={ 10 }^{ k },where\quad m<k<m+1\quad \\ k=log2^{1000}=1000log2\approx 301.02999...,so\quad m=[1000\log { 2 } ]=301\\ \therefore { 2 }^{ 1000 }\quad contains\quad 302\quad digits$ .

$Let\quad \\ { 2 }^{ 1000 }=\quad x\\ taking\quad log\quad both\quad the\quad sides\\ (1000)\log { 2=\log { x } } \\ 301.029=\log { x } \\ by\quad this\quad we\quad can\quad say\quad that\quad x\quad has\quad 302\quad digits$

log 2^1000 = 1000 log 2 =1000(0.3010)=301=302 digits