Question

If $\frac{2}{{\log }_{b}x}=\frac{1}{{\log }_{a}x}+\frac{1}{{\log }_{c}x}$ then prove that $a,b,c$ are in $GP.$ Where $(a,b,c$ $\&$ $x>0)$ $\&$ $\ne 1)$

Answer 1

$\frac{2}{{log}_{b}x}=\frac{1}{{log}_{a}x}+\frac{1}{{log}_{c}x}$

$2{log}_{x}b={log}_{x}a+{log}_{x}c$

$2{log}_{x}b={log}_{x}ac$

${log}_x{b}^2={log}_{x}ac$

$b^2=ac$

Hence $a,b,c$ are in $GP$