Question

$7log\left(\frac{32}{15}\right)+5log\left(\frac{25}{24}\right)+3log\left(\frac{81}{160}\right)$ is equal to

• A. 5$log2$
• B. 1
• C. $log2$
• D. $log3$

Answer 1

The correct option is A

5$log2$

Given expression becomes:
$7\left(5log\right(2)-log(3)-log(5\left)\right)+5\left(2log\right(5)-log(3)-3log(2\left)\right)+3\left(4log\right(3)-log(5)-5log(2\left)\right)=35log\left(2\right)-30log\left(2\right)+12log\left(3\right)-5log\left(3\right)-7log\left(3\right)+10log\left(5\right)-7log\left(5\right)-3log\left(5\right)=5log\left(2\right)$