Question

$7\log \left(\frac{16}{15}\right)+5\log \left(\frac{25}{24}\right)+3\log \left(\frac{81}{80}\right)$ is equal to

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

Answer 1

The correct option is C

$\log 2$

Explanation for the correct option:

Use logarithmic properties to simplify the expression:

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

$\begin{array}{rcl}7\log \left(\frac{16}{15}\right)+5\log \left(\frac{25}{24}\right)+3\log \left(\frac{81}{80}\right)& =& \log {\left(\frac{16}{15}\right)}^7{\left(\frac{25}{24}\right)}^5{\left(\frac{81}{80}\right)}^3\left[\because \log \left(a\right)+\log \left(b\right)=\log \left(\mathrm{ab}\right)\right]\\ & =& \log \left(\frac{2^{28}}{3^7\times 5^7}·\frac{5^{10}}{2^{15}\times 3^5}·\frac{3^{12}}{2^{12}\times 5^3}\right)\\ & =& \log 2\end{array}$

Hence, option (C) is the correct answer.