Question

If ${\log }_{30}3=a,$ and ${\log }_{30}5=b$, then value of ${\log }_{30}8$ is

• A. $ab$
• B. $3(1-a-b)$
• C. $a+b$
• D. $\frac{1}{a+b}$

Answer 1

Answer: (B)

$3(1-a-b)$

${\log }_{30}3=a$ and ${\log }_{30}5=b$

To find ${\log }_{30}8$, we will use ${\log }_{30}30=1$

$⟹{\log }_{30}30=1⟹{\log }_{30}(3\times 5\times 2)=1⟹{\log }_{30}3+{\log }_{30}5+{\log }_{30}2=1⟹a+b+{\log }_{30}2=1⟹{\log }_{30}2=(1-a-b)\therefore {\log }_{30}8={\log }_{30}{2}^3=3{\log }_{30}2=3(1-a-b)$