Question

Which of the following when simplified, reduces to an integer?

• A. $\frac{2\log 6}{\log 12+\log 3}$
• B. $\frac{\log 32}{\log 4}$
• C. $\frac{{\log }_{5}16-{\log }_{5}4}{{\log }_{5}128}$
• D. ${\log }_{\frac{1}{4}}{\left(\frac{1}{16}\right)}^2$

Answer 1

Answer: (A), (D)

$\frac{2\log 6}{\log 12+\log 3}$
${\log }_{\frac{1}{4}}{\left(\frac{1}{16}\right)}^2$

Also ${\log }_{a}a=1$

1. $\frac{2\log 6}{\log 12+\log 3}$$=\frac{\log 36}{\log 12+\log 3}$

$=\frac{\log 36}{\log 36}=1$

2. $\frac{\log 32}{\log 4}$$=\frac{\log 2^5}{\log 2^2}$$=\frac{5}{2}$

3. $\frac{{\log }_{5}16-{\log }_{5}4}{{\log }_{5}128}$$=\frac{{\log }_{5}4}{{\log }_{5}128}$

$=\frac{{\log }_{5}4}{{\log }_{5}128}$$=\frac{2}{7}$

4. ${\log }_{\frac{1}{4}}{\left(\frac{1}{16}\right)}^2$$={\log }_{\frac{1}{4}}{\left(\frac{1}{4}\right)}^4=4$

Hence, options A and D.