Factorial Log

$\frac{1}{\log_{2}{100!}}=\frac{1}{\frac{\log{100!}}{log{2}}}=\frac{\log {2}}{\log {100!}}$

So

$\frac{1}{\log_{2}{100!}}+\frac{1}{\log_{3}{100!}}+\frac{1}{log_{4}{100!}}+...+\frac{1}{\log_{100}{100!}}=\frac{\log {2}}{\log {100!}}+\frac{\log {3}}{\log {100!}}+\frac{\log {4}}{\log {100!}}+...+\frac{\log {100}}{\log {100!}}$

By the Properties of Logarithms , $\log A+ \log B=\log AB$

$\frac{\log {2}}{\log {100!}}+\frac{\log {3}}{\log {100!}}+\frac{\log {4}}{\log {100!}}+...+\frac{\log {100}}{\log {100!}}=\frac{\log{100!}}{\log{100!}}=\boxed{1}$

Remember that $\frac{1}{\log_2 {100!}}=\log_{100!}2$ by the base change formula $\log_ba=\frac{\log_xa}{\log_xb}$ . Apply this idea all over. You will end up in this $\log_{100!}2+\log_{100!}3+\cdots+\log_{100!}100=\log_{100!}{(2\cdot3\cdots 100)}=\log_{100!}{100!}=1$

You can also remember the formula $\log_b a = \frac{1}{\log_a b}$ Which in turn makes this question into $\log_{100!}{2}+\log_{100!}{3}+\log_{100!}{4}+\ldots+\log_{100!}{100}$ $=log_{100!}{(2*3*4\ldots*100)}$ $=log_{100!}{100!}$ $=1$

$\frac{1}{\log_{2}{100!}}+\frac{1}{\log_{3}{100!}}+\frac{1}{\log_{4}{100!}}+\ldots+\frac{1}{\log_{100}{100!}}= \ ?$

$\frac{\log_{2}{2}}{\log_{2}{100!}}+\frac{\log_{3}{3}}{\log_{3}{100!}}+\frac{\log_{4}{4}}{\log_{4}{100!}}+\ldots+\frac{\log_{100}{100}}{\log_{100}{100!}}$

$\text{Because} \quad \color{#302B94}{\frac{\log_{a}{b}}{\log_{a}{c}}=\log_{c}{b}}$

$\large\log_{100!}{2}+\log_{100!}{3}+\log_{100!}{4}+\cdots+\log_{100!}{100}$

$\text{Using the} \, \color{#3D99F6}{\text{Law of Logarithm,}} \, \color{#302B94}{\log MN=\log M+\log N}$

$\large\log_{100!}{\color{#D61F06}{\big(2*3*4*\cdots *100)}}$

$\large\log_{100!}{\color{#D61F06}{100!}}$

$\large\implies\color{#302B94}{\boxed{1}}$

Good Solution

summary log 100!/100! = 1

log y base x gives us log y / logx ,we here have the 1/ log 100! base n , thus we get 1/log100! [ log2 + log3 + ..... log100 ] , since logarithms are added then resultant logarithm is multiplication of the terms , then we get log100!/log100! = 1

1/log2(100!) + 1/log3(100!) + 1/log4(100!) +............+1/log100(100!) = log(2)/log(100!) +log(3)/log(100!) + log(4)/log(100!) +..........+ log(100)/log(100!) =(log(2) + log(3) + log(4) +.........+ log(100))/log(100!) As log(A) + log(B) = log(AB); =log(2 3 4 ..... 100)/log(100!) =log(100!)/log(100!) =1

convert to natural log, log(base2)100!=ln2/ln100! and so on! so denominator becomes ln100! and numerator is ln2+ln3+ln4+.....+ln100=ln(2 3 4...*100)=ln 100 factorial GONDEN RULE: log(base A)B = 1/log(base B)A