Question

Find the value of ${\log }_{10}\frac{{64}^{2.1}\times {81}^{4.2}}{{49}^{3.4}}$ using log table

• A. $2.1\times 6\times .303+4.2\times 2\times .854-3.4\times 2\times .745$
• B. $2.1\times .303+4.2\times .954-3.4\times .845$
• C. $2.1\times 6\times .303-4.2\times 2\times .954+3.4\times 2\times .845$
• D. $2.1\times 6\times .303+4.2\times 2\times .954-3.4\times 2\times .845$

Answer 1

The correct option is D $2.1\times 6\times .303+4.2\times 2\times .954-3.4\times 2\times .845$

${\log }_{10}{64}^{2.1}+{\log }_{10}{81}^{4.2}-{\log }_{10}{49}^{3.4}$

$=2.1{\log }_{10}64+4.2{\log }_{10}81-3.4{\log }_{10}49$

$=2.1{\log }_{10}{2}^6+4.2{\log }_{10}{3}^4-3.4{\log }_{10}{7}^2$

$=2.1\times 6\times \left(0.303\right)+4.2\times 4\times \left(0.477\right)-3.4\times 2\times \left(0.845\right)$

$=2.1\times 6\times \left(0.303\right)+4.2\times 2\times \left(0.954\right)-3.4\times 2\times \left(0.845\right)$