Question

The value of $4\log \frac{24}{25}-16\log \frac{9}{10}+7\log \frac{81}{80}$

• A. $\log 9$
• B. $\log 7$
• C. $\log 5$
• D. $\log 3$

Answer 1

The correct option is C

$\log 5$

Solution:

$$\begin{gathered} 4\log \frac{24}{25}-16\log \frac{9}{10}+7\log \frac{81}{80} \\ =\log {\left(\frac{24}{25}\right)}^4-\log {\left(\frac{9}{10}\right)}^{16}+\log {\left(\frac{81}{80}\right)}^7\left[\mathbf{\because }\mathbf{m}\mathbf{}\mathbf{log}\mathbf{}\mathbf{a}\mathbf{=}\mathbf{log}\mathbf{}{\mathbf{a}}^{\mathbf{m}}\right] \\ =\log {\left(\frac{2^3\times 3}{5^2}\right)}^4-\log {\left(\frac{3^2}{2\times 5}\right)}^{16}+\log {\left(\frac{3^4}{2^4\times 5}\right)}^7 \\ =\log \frac{\left[{\left({\displaystyle \frac{2^3\times 3}{5^2}}\right)}^4\times {\left(\frac{3^4}{2^4\times 5}\right)}^7\right]}{{\left(\frac{3^2}{2\times 5}\right)}^{16}}\left[\mathbf{\because }\mathbf{log}\mathbf{}\mathbf{a}\mathbf{-}\mathbf{log}\mathbf{}\mathbf{b}\mathbf{+}\mathbf{log}\mathbf{}\mathbf{c}\mathbf{=}\mathbf{log}\mathbf{\left(}\frac{\mathbf{a}\mathbf{\times }\mathbf{c}}{\mathbf{b}}\mathbf{\right)}\right] \\ =\log \left(\frac{{\displaystyle \frac{2^{12}\times 3^4\times 3^{28}}{5^8\times 2^{28}\times 5^7}}}{{\displaystyle \frac{3^{32}}{2^{16}\times 5^{16}}}}\right)\left[\because {\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{mn}}\right] \\ =\log \left(2^{12+16-28}\times 3^{4+28-32}\times 5^{16-8-7}\right)\left[\mathbf{\because }\frac{{\mathbf{a}}^{\mathbf{m}}\mathbf{\times }{\mathbf{a}}^{\mathbf{n}}}{{\mathbf{a}}^{\mathbf{l}}}\mathbf{=}{\mathbf{a}}^{\mathbf{m}\mathbf{+}\mathbf{n}\mathbf{-}\mathbf{l}}\right] \\ =\mathbf{log}\mathbf{5} \end{gathered}$$

Final answer: Hence, option C is the correct answer