Question

The value of, log102+16log10(16/15)+12log10(25/24)+7log10(81/80)log10⁡2+16log10⁡(16/15)+12log10⁡(25/24)+7log10⁡(81/80) is

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

${\log }_{10}2+16{\log }_{10}\left(16/15\right)+12{\log }_{10}\left(25/24\right)+7{\log }_{10}\left(81/80\right)$

$={\log }_{10}[2\times {\left(\frac{16}{15}\right)}^{16}\times {\left(\frac{25}{24}\right)}^{12}\times {\left(\frac{81}{80}\right)}^7]$

$={\log }_{10}[2\times \left(\frac{(2^4{)}^{16}}{3^{16}\times 5^{16}}\right)\times \left(\frac{(5^2{)}^{12}}{(2^3{)}^{12}\times (3{)}^{12}}\right)\times \left(\frac{(3^7{)}^7}{(2^4{)}^7\times (5{)}^7}\right)]$

$={\log }_{10}[\left(\frac{2\times 2^{64}}{2^{36}\times 2^{28}}\right)\times \left(\frac{3^{28}}{3^{16}\times 3^{12}}\right)\times \left(\frac{5^{24}}{5^{16}\times 5^7}\right)]$

$={\log }_{10}[\left(\frac{2^{65}}{2^{64}}\right)\times \left(\frac{3^{28}}{3^{28}}\right)\times \left(\frac{5^{24}}{5^{23}}\right)]$

$={\log }_{10}[{\left(2\right)}^{65-64}\times {\left(5\right)}^{24-23}]$

$={\log }_{10}[2\times 5]$

$={\log }_{10}10=1$