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 $27{\log }_{b}a$ is

Answer 1

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, $27{\log }_{b}a=27{\log }_{2^3}{2}^2=\frac{2}{3}\times 27=18$