Question

The value of $7{\log }_a\frac{16}{15}+5{\log }_a\frac{25}{24}+3{\log }_a\frac{81}{80}$ is

• A. ${\log }_{a}3$
• B. ${\log }_{a}1$
• C. ${\log }_{a}2$
• D. ${\log }_{a}5$

Answer 1

Answer: (C)

${\log }_{a}2$

$7{\log }_a\frac{16}{15}+5{\log }_a\frac{25}{24}+3{\log }_a\frac{81}{80}$

$={\log }_a{\left(\frac{16}{15}\right)}^7+{\log }_a{\left(\frac{25}{24}\right)}^5+{\log }_a{\left(\frac{81}{80}\right)}^3$

$={\log }_a{\left(\frac{16}{15}\right)}^7\times {\left(\frac{25}{24}\right)}^5\times {\left(\frac{81}{80}\right)}^3$

$={\log }_a[\frac{4^7\times 4^7}{3^7\times 5^7}\times \frac{5^{10}}{6^5\times 4^5}\times \frac{3^{12}}{4^6\times 5^3}]$

$={\log }_a[4^{7+7-5-6}\times 5^{10-7-3}\times \frac{3^{12-7}}{6^5}]$

$={\log }_a[4^3\times 5^0\times \frac{3^5}{3^5\times 2^5}]$

$={\log }_a[2^6\times \frac{3^5}{3^5\times 2^5}]$

$={\log }_a[2^{6-5}\times 3^{5-5}]$

$={\log }_{a}2$