Question

Simplify the following.
(i) ${\log }_{10}3+{\log }_{10}3$
(ii) ${\log }_{25}35-{\log }_{25}10$
(iii) ${\log }_{7}21+{\log }_{7}77+{\log }_{7}88-{\log }_{7}121-{\log }_{7}24$
(iv) ${\log }_{8}16+{\log }_{8}52-\frac{1}{{\log }_{13}8}$
(v) $5{\log }_{10}2+2{\log }_{10}3-6{\log }_{64}4$
(vi) ${\log }_{10}8+{\log }_{10}5-{\log }_{10}4$

Answer 1

(i) $2{\log }_{10}3={\log }_{10}{3}^2={\log }_{10}9$

(ii) ${\log }_{25}\frac{35}{10}={\log }_{5^2}\frac{7}{2}=2{\log }_5\left(3.5\right)$

(iii) ${\log }_7\left\{\frac{21\times 77\times 88}{121\times 24}\right\}{\log }_{7}49=2$

(iv) ${\log }_{8}16+{\log }_{8}52-{\log }_{8}13={\log }_8\left\{\frac{16\times 52}{13}\right\}$

$={\log }_{8}64={\log }_8{8}^2=2$

(v) ${\log }_{10}{2}^5+{\log }_{10}{3}^2-6{\log }_{2^6}4$

$={\log }_{10}{2}^5\times 3^2-6\times \frac{1}{6}{\log }_{2}4$

$={\log }_{10}(2^5\times 3^2)-2{\log }_{10}10$

$={\log }_{10}\left(\frac{32\times 9}{100}\right)={\log }_{10}\left(2.88\right)$

(vi) ${\log }_{10}\left(\frac{8\times 5}{4}\right)={\log }_{10}10=1$