Question

Find the value of $7\log \left(\frac{16}{15}\right)+5\log \left(\frac{25}{24}\right)+3\log \left(\frac{81}{80}\right)$

• A. $\log 2$
• B. $\log 4$
• C. $\log 9$
• D. $\log 15$

Answer 1

The correct option is A $\log 2$

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

$=\log {\left(\frac{81}{80}\right)}^3+\log {\left(\frac{25}{24}\right)}^5+\log {\left(\frac{16}{15}\right)}^7[\because m\log a=\log a^m]$

$=\log ({\left(\frac{81}{80}\right)}^3\times {\left(\frac{25}{24}\right)}^5\times {\left(\frac{16}{15}\right)}^7)[\because \log a+\log b+\log c=\log (abc\left)\right]$

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

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