Question

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

• A. $\log 2$
• B. $\log 3$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

$\log 2$

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

$=\log ({\left(\frac{3^4}{2^4\times 5}\right)}^3\times {\left(\frac{5^2}{2^3\times 3}\right)}^5\times {\left(\frac{2^4}{3\times 5}\right)}^7)$
$=\log \left(\frac{3^{12}\times 5^{10}\times 2^{28}}{2^{12}\times 5^3\times 2^{15}\times 3^5\times 3^7\times 5^7}\right)=\log 2$