Question

If anti ${\log }_{10}\left(0.3678\right)=2.3324$ then ${\log }_{10}233.2$ is equal to

• A. 367.8
• B. 36.78
• C. 3.3678
• D. 2.3678

Answer 1

Answer: (D)

2.3678

$antilo{g}_{10}\left(0.3678\right)=2.3324$

$\therefore \log \left(2.3324\right)=0.3678$

${\log }_{10}233.2={\log }_{10}(2.332\times 100)$

$={\log }_{10}\left(2.332\right)+\log \left({10}^2\right)\dots (\log a+\log b=\log (ab\left)\right)$

$={\log }_{10}\left(2.332\right)+2\log \left(10\right)\dots (n{\log }_{a}b={\log }_a{b}^n)$

As we know, $lo{g}_{a}a=1$

${\log }_{10}233.2=2+0.3678=2.3678$