Question

If x, y, z are in Gp and

$\sf \: {a}^{x} = {b}^{y} = {c}^{z} \: prove \: that \: log_{b}(a) = log_{c}(b)$
​

Answer 1

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

Given that, x, y, z are in GP.

So, common ratio between consecutive terms is same.

Thus,

$\rm \implies\:\dfrac{y}{x} = \dfrac{z}{y} - - - (1)$

Further given that

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

On taking log on both sides, we get

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

$\rm :\longmapsto\:x \: loga \: = \: y \: logb \: = \: z \: logc$

So, Taking first and second member,

$\rm :\longmapsto\:x \: loga \: = \: y \: logb \:$

$\rm :\longmapsto\:\dfrac{y}{x} = \dfrac{loga}{logb} - - - - (2)$

Taking second and third member, we get

$\rm :\longmapsto\: y \: logb \: = \: z \: logc$

$\rm :\longmapsto\:\dfrac{z}{y} = \dfrac{logb}{logc} - - - - (3)$

So, equating equations (1), (2) and (3), we get

$\rm :\longmapsto\:\dfrac{loga}{logb} = \dfrac{logb}{logc}$

$\bf\implies \: log_{b}(a) = log_{c}(b)$

Hence, Proved

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Formula Used :-

$\boxed{ \tt{ \: y \: logx = log {x}^{y} \: }}$

$\boxed{ \tt{ \: \frac{logx}{logy} = log_{y}(x) \: }}$

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Additional Information :-

$\boxed{ \tt{ \: log1 = 0 \: }}$

$\boxed{ \tt{ \: {e}^{logx} \: = \: x \: }}$

$\boxed{ \tt{ \: {e}^{y \: logx} \: = \: {x}^{y} \: }}$

$\boxed{ \tt{ \: {a}^{ log_{a}(x) } = x \: }}$

$\boxed{ \tt{ \: {a}^{y log_{a}(x) } = {x}^{y} \: }}$

$\boxed{ \tt{ \: log_{x}(x) = 1 \: }}$