Question

if p1//x = q1/y = r1/z and pqr = 1 . Prove that x+y+z = 0

Answer 1

Appropriate Question :-

If

$\rm :\longmapsto\: {\bigg[p\bigg]}^{\dfrac{1}{x} } = {\bigg[q\bigg]}^{\dfrac{1}{y} } = {\bigg[r\bigg]}^{\dfrac{1}{z} } \: and \: pqr = 1$

Prove that x + y + z = 0

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

Given that,

$\rm :\longmapsto\: {\bigg[p\bigg]}^{\dfrac{1}{x} } = {\bigg[q\bigg]}^{\dfrac{1}{y} } = {\bigg[r\bigg]}^{\dfrac{1}{z} }$

Let assume that

$\rm :\longmapsto\: {\bigg[p\bigg]}^{\dfrac{1}{x} } = {\bigg[q\bigg]}^{\dfrac{1}{y} } = {\bigg[r\bigg]}^{\dfrac{1}{z} } = k$

So,

$\rm :\longmapsto\: {\bigg[p\bigg]}^{\dfrac{1}{x} } = k \: \bf\implies \:p = {(k)}^{x}$

Also,

$\rm :\longmapsto\: {\bigg[q\bigg]}^{\dfrac{1}{y} } = k \: \bf\implies \:q = {(k)}^{y}$

Also,

$\rm :\longmapsto\: {\bigg[r\bigg]}^{\dfrac{1}{z} } = k \: \bf\implies \:r = {(k)}^{z}$

Further, given that

$\rm :\longmapsto\:pqr = 1$

On substituting the values of p, q and r, we get

$\rm :\longmapsto\: {k}^{x} \times {k}^{y} \times {k}^{z} = 1$

We know,

$\boxed{ \tt{ \: {a}^{m} \times {a}^{n} = {a}^{m + n}}}$

So, using this, we get

$\rm :\longmapsto\: {k}^{x + y + z} = 1$

$\rm :\longmapsto\: {k}^{x + y + z} = {k}^{0}$

$\bf\implies \:x + y + z = 0$

Hence, Proved

More to know :-

$\boxed{ \tt{ \: {a}^{m} \times {a}^{n} = {a}^{m + n}}}$

$\boxed{ \tt{ \: {a}^{m} \div {a}^{n} = {a}^{m - n}}}$

$\boxed{ \tt{ \: {( {a}^{n} )}^{m} = {a}^{mn}}}$

$\boxed{ \tt{ \: {a}^{ - n} = \frac{1}{ {a}^{n} }}}$

$\boxed{ \tt{ \: {x}^{0} = 1}}$

Answer 2

Answer:

Appropriate Question :-

If

$\rm :\longmapsto\: {\bigg[p\bigg]}^{\dfrac{1}{x} } = {\bigg[q\bigg]}^{\dfrac{1}{y} } = {\bigg[r\bigg]}^{\dfrac{1}{z} } \: and \: pqr = 1$

Prove that x + y + z = 0

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

Given that,

$\rm :\longmapsto\: {\bigg[p\bigg]}^{\dfrac{1}{x} } = {\bigg[q\bigg]}^{\dfrac{1}{y} } = {\bigg[r\bigg]}^{\dfrac{1}{z} }$

Let assume that

$\rm :\longmapsto\: {\bigg[p\bigg]}^{\dfrac{1}{x} } = {\bigg[q\bigg]}^{\dfrac{1}{y} } = {\bigg[r\bigg]}^{\dfrac{1}{z} } = k$

So,

$\rm :\longmapsto\: {\bigg[p\bigg]}^{\dfrac{1}{x} } = k \: \bf\implies \:p = {(k)}^{x}$

Also,

$\rm :\longmapsto\: {\bigg[q\bigg]}^{\dfrac{1}{y} } = k \: \bf\implies \:q = {(k)}^{y}$

Also,

$\rm :\longmapsto\: {\bigg[r\bigg]}^{\dfrac{1}{z} } = k \: \bf\implies \:r = {(k)}^{z}$

Further, given that

$\rm :\longmapsto\:pqr = 1$

On substituting the values of p, q and r, we get

$\rm :\longmapsto\: {k}^{x} \times {k}^{y} \times {k}^{z} = 1$

We know,

$\boxed{ \tt{ \: {a}^{m} \times {a}^{n} = {a}^{m + n}}}$

So, using this, we get

$\rm :\longmapsto\: {k}^{x + y + z} = 1$

$\rm :\longmapsto\: {k}^{x + y + z} = {k}^{0}$

$\bf\implies \:x + y + z = 0$

Hence, Proved

More to know :-

$\boxed{ \tt{ \: {a}^{m} \times {a}^{n} = {a}^{m + n}}}$

$\boxed{ \tt{ \: {a}^{m} \div {a}^{n} = {a}^{m - n}}}$

$\boxed{ \tt{ \: {( {a}^{n} )}^{m} = {a}^{mn}}}$

$\boxed{ \tt{ \: {a}^{ - n} = \frac{1}{ {a}^{n} }}}$

$\boxed{ \tt{ \: {x}^{0} = 1}}$