Question

The value of ${\log }_4(1+\frac{1}{4})+{\log }_4(1+\frac{1}{5})+{\log }_4(1+\frac{1}{6})+...+{\log }_4(1+\frac{1}{255})=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (D)

$3$

$={\log }_4(\frac{5}{4})+{\log }_4(\frac{6}{5})+{\log }_4(\frac{7}{6})+...+{\log }_4(\frac{256}{255})$

$={\log }_{4}5-{\log }_{4}4+{\log }_{4}6-{\log }_{4}5+{\log }_{4}7-{\log }_{4}6+...+{\log }_{4}256-{\log }_{4}255$ (since, $\log \left(\frac{a}{b}\right)=\log a-\log b$)

$=-{\log }_{4}4+{\log }_{4}256={\log }_4{4}^4-{\log }_{4}4$

$=4{\log }_{4}4-{\log }_{4}4$ (since, ${\log }_b{x}^a=a{\log }_{b}x$)

$4-1=3$ (since, ${\log }_{b}b=1$)