Question

Find the value of the expression $\frac{2}{{\log }_4{\left(2000\right)}^6}+\frac{3}{{\log }_5{\left(2000\right)}^6}$.

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $\frac{1}{12}$

Answer 1

The correct option is C $\frac{1}{6}$

$\frac{2}{{\log }_4(2000{)}^6}+\frac{3}{{\log }_5(2000{)}^6}$

let $x={2000}^6$

$\frac{2}{{\log }_4\left(x\right)}+\frac{3}{{\log }_5\left(x\right)}$

use the change of base formula

$\left[{\log }_b\right(a)=\frac{{\log }_{10}\left(a\right)}{{\log }_{10}\left(b\right)}]$

$=\frac{2}{\frac{{\log }_{10}\left(x\right)}{{\log }_{10}\left(4\right)}}+\frac{3}{\frac{{\log }_{10}\left(x\right)}{{\log }_{10}\left(5\right)}}$

$=\frac{2{\log }_{10}\left(4\right)}{{\log }_{10}\left(x\right)}+\frac{3{\log }_{10}\left(3\right)}{{\log }_{10}\left(x\right)}$

$=\frac{2{\log }_{10}\left(4\right)+3{\log }_{10}\left(5\right)}{{\log }_{10}\left(x\right)}$

$=\frac{{\log }_{10}\left(4^2\right)+{\log }_{10}\left(5^3\right)}{{\log }_{10}\left(x\right)}$

$=\frac{{\log }_{10}\left(16\right)+{\log }_{10}\left(125\right)}{{\log }_{10}\left(x\right)}$

$=\frac{{\log }_{10}(16\times 125)}{{\log }_{10}\left(x\right)}$

Now put the value of $x={2000}^6$

$=\frac{{\log }_{10}\left(2000\right)}{{\log }_{10}\left({2000}^6\right)}$

$=\frac{{\log }_{10}\left(2000\right)}{6{\log }_{10}\left(2000\right)}$

$=\frac{1}{6}$