Question

if 3^x = 5^y = 45^z , prove that x= 2yz/y-z

Answer 1

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

$\red{\rm :\longmapsto\: {3}^{x} = {5}^{y} = {45}^{z}}$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:\boxed{ \rm{ x = \frac{2yz}{y - z}}}$

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

Given that

$\red{\rm :\longmapsto\: {3}^{x} = {5}^{y} = {45}^{z}}$

Let us assume that

$\red{\rm :\longmapsto\: {3}^{x} = {5}^{y} = {45}^{z} = k}$

$\rm :\implies\: {3}^{x} = k$

$\rm :\implies\: {5}^{y} = k$

$\rm :\implies\: {45}^{z} = k$

We know,

$\boxed{ \rm{If \: \: \: {x}^{y} = z \: \rm \implies\:x = {\bigg(z\bigg) }^{\dfrac{1}{y} } }}$

So, using this identity, we get

$\rm :\longmapsto\:3 = {\bigg(k \bigg) }^{\dfrac{1}{x} }$

$\rm :\longmapsto\:5 = {\bigg(k \bigg) }^{\dfrac{1}{y} }$

and

$\rm :\longmapsto\:45 = {\bigg(k \bigg) }^{\dfrac{1}{z} }$

can be rewritten as

$\rm :\longmapsto\:3 \times 3 \times 5 = {\bigg(k \bigg) }^{\dfrac{1}{z} }$

$\rm :\longmapsto\: {3}^{2} \times 5 = {\bigg(k \bigg) }^{\dfrac{1}{z} }$

$\rm :\longmapsto\:{\bigg(k\bigg) }^{\dfrac{2}{x} } \times {\bigg(k\bigg) }^{\dfrac{1}{y} } = {\bigg(k \bigg) }^{\dfrac{1}{z} }$

We know,

$\boxed{ \rm{ {k}^{x} \times {k}^{y} = {k}^{x + y}}}$

So, using this identity, we get

$\rm :\longmapsto\:{\bigg(k\bigg) }^{\dfrac{2}{x} + \dfrac{1}{y} }= {\bigg(k \bigg) }^{\dfrac{1}{z} }$

We know,

$\boxed{ \rm{ If \: \: {k}^{x} = {k}^{y} \rm \implies\:x = y}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{2}{x} + \dfrac{1}{y} = \dfrac{1}{z}$

$\rm :\longmapsto\:\dfrac{2}{x} = \dfrac{1}{z} - \dfrac{1}{y}$

$\rm :\longmapsto\:\dfrac{2}{x} = \dfrac{y - z}{zy}$

$\rm :\longmapsto\:\dfrac{1}{x} = \dfrac{y - z}{2zy}$

$\rm :\longmapsto\:x = \dfrac{2yz}{y - z}$

Hence, Proved