Question

if a^x = b^y = c^z and x, y, z are im G. P then,

please answer me quickly!!​

Answer 1

Question

$\sf \: If \: {a}^{x} = {b}^{y} = {c}^{z} \: and \:x, \: y , \: z \: are \: in \: GP \: then$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: \sf \: \: {a}^{x} = {b}^{y} = {c}^{z} \:$

Let assume that

$\rm :\longmapsto\: \sf \: \: {a}^{x} = {b}^{y} = {c}^{z} \: = k$

$\red{\bf :\longmapsto\: {a}^{x} = k\rm \: \: \: \implies\:x = log_{a}(k) - - - (1)}$

$\: \: \: \: \: \: \: \red{\bigg \{ \because \: {a}^{b} = c \: \: then \: \: b = log_{a}(c) \bigg \}}$

$\green{\bf :\longmapsto\: {b}^{y} = k\rm \: \: \: \implies \: y = log_{b}(k) - - - (2)}$

$\blue{\bf :\longmapsto\: {c}^{z} = k\rm \: \: \: \implies\:z = log_{c}(k) - - - (3)}$

Again,

It is given that

$\bf :\longmapsto\:x, \: y, \: z \: are \: in \: GP$

$\rm :\longmapsto\:\dfrac{y}{x} = \dfrac{z}{y}$

$\rm :\longmapsto\: {y}^{2} = xz$

On substituting the values of x, y, z from (1), (2) and (3),

$\rm :\longmapsto\: {( log_{b}(k)) }^{2} = log_{a}(k) \times log_{c}(k)$

$\rm :\longmapsto\: {\bigg(\dfrac{logk}{logb} \bigg) }^{2} = \dfrac{logk}{loga} \times \dfrac{logk}{logc}$

$\rm :\longmapsto\: {\bigg(\dfrac{1}{logb} \bigg) }^{2} = \dfrac{1}{loga} \times \dfrac{1}{logc}$

$\bf\implies \: {(logb)}^{2} = loga \: \times \: logc$

$\bf\implies \:loga, \: logb, \: logc \: are \: in \: GP$