Question

If $a,b,c$ are in G.P then${\log }_{a}x,{\log }_{b}x,{\log }_{c}x$ are in

Answer 1

Identify the series of ${\log }_{a}x,{\log }_{b}x,{\log }_{c}x$.

Given: $a,b,c$ are in G.P.

So, $b^2=ac$

Taking $\log$ on both sides :

$\Rightarrow$ $\log \left(b^2\right)=\log \left(ac\right)$ ; $\left[\log \left(m^n\right)=n\log \left(m\right)\right]$

$\Rightarrow$ $2\log \left(b\right)=\log \left(a\right)+\log \left(c\right)$

Divide by $\log \left(x\right)$ on both sides.

$\Rightarrow$ $2\frac{\log \left(b\right)}{\log \left(x\right)}=\frac{\log \left(a\right)}{\log \left(x\right)}+\frac{\log \left(c\right)}{\log \left(x\right)}$ ; $\left[{\log }_m\left(n\right)=\frac{\log \left(n\right)}{\log \left(m\right)}\right]$

$\Rightarrow$ $2{\log }_x\left(b\right)={\log }_x\left(a\right)+{\log }_x\left(c\right)$

$\Rightarrow$ $\frac{2}{{\log }_b\left(x\right)}=\frac{1}{{\log }_a\left(x\right)}+\frac{1}{{\log }_c\left(x\right)}$ ; $\left[{\log }_m\left(n\right)=\frac{1}{{\log }_n\left(m\right)}\right]$

We know, if three terms $a,b,c$ are in A.P then we can write $2b=a+c$

$\Rightarrow$$2\left(\frac{1}{{\log }_{b}x}\right)=\frac{1}{{\log }_{a}x}+\frac{1}{{\log }_{c}x}$

Thus, $\frac{1}{{\log }_a\left(x\right)},\frac{1}{{\log }_b\left(x\right)},\frac{1}{{\log }_c\left(x\right)}$ are in A.P.

Since if $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P then $a,b,c$are in H.P.

Hence, ${\log }_{a}x,{\log }_{b}x,{\log }_{c}x$ are in H.P.