Question

If $x=1+\log 2-\log 5$, $y=2\log 3$, $z=\log 3m-\log 5$ and $x+y=2z$, then the value of $m$ is equal to

• A. $12$
• B. $20$
• C. $10$
• D. $24$

Answer 1

The correct option is C $10$

$x=1+\log 2-\log 5\text{Assuming base 10}x={\log }_{10}10+{\log }_{10}2-{\log }_{10}5={\log }_{10}\frac{20}{5}={\log }_{10}4y=2{\log }_{10}3={\log }_{10}9z={\log }_{10}3m-{\log }_{10}5={\log }_{10}\frac{3m}{5}x+y=2zSo...{\log }_{10}4+{\log }_{10}9=2{\log }_{10}\frac{3m}{5}{\log }_{10}(4\times 9)={\log }_{10}{\left(3m\right)}^2-{\log }_{10}{5}^2\log 36+\log 25=\log {9m}^2\log (36\times 25)=\log {9m}^2\text{Removing log}9m^2=36\times 25m^2=4\times 25m=10$