Question

The arithmetic mean of two numbers $a$ and $b$ $(a<b)$ is $6.$ If the geometric mean $G$ and harmonic mean $H$ of the two numbers satisfy the relation $G^2+3H=48,$ then the value of ${\log }_{b}a$ is

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

Answer 1

The correct option is C $\frac{2}{3}$

Given, $\frac{a+b}{2}=6$
$\Rightarrow a+b=12\cdots \left(1\right)$

$G^2+3H=48$
$\Rightarrow ab+\frac{3\times 2ab}{a+b}=48$
$\Rightarrow ab+\frac{ab}{2}=48\left[\text{From (1)}\right]$
$\Rightarrow ab=32\cdots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get
$a=4,b=8$ as $a<b$
Hence, ${\log }_{b}a={\log }_{2^3}{2}^2=\frac{2}{3}$