Question

if 3^p=4^q=12^-r, then find the value of (1/p)+(1/q)+(1/r)

Answer 1

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

Given expression is

$\rm :\longmapsto\: {3}^{p} = {4}^{q} = {12}^{ - r}$

Let we assume that

$\rm :\longmapsto\: {3}^{p} = {4}^{q} = {12}^{ - r} = k$

We know,

$\boxed{\tt{ {x}^{m} = y \: \: \rm \implies\:3{\bigg(y \bigg) }^{\dfrac{1}{m} }}}$

So, using this Law of exponents, we get

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

$\red{\rm :\longmapsto\: {4}^{q} = k \: \: \rm \implies\:4 = {\bigg(k \bigg) }^{\dfrac{1}{q} } \: }$

$\red{\rm :\longmapsto\: {12}^{ - r} = k \: \: \rm \implies\:12 = {\bigg(k \bigg) }^{ - \dfrac{1}{r} } \: }$

Now,

$\rm :\longmapsto\:12 = {\bigg(k \bigg) }^{ - \dfrac{1}{r} }$

can be rewritten as

$\rm :\longmapsto\:4 \times 3 = {\bigg(k \bigg) }^{ - \dfrac{1}{r} }$

$\rm :\longmapsto\:{\bigg(k \bigg) }^{\dfrac{1}{p} } \times {\bigg(k \bigg) }^{\dfrac{1}{q} } = {\bigg(k \bigg) }^{ - \dfrac{1}{r} }$

We know,

$\boxed{\tt{ \: \: {x}^{p} \: \times \: {x}^{q} \: = \: {x}^{p \: + \: q} \: \: }}$

So, using this, we get

$\rm :\longmapsto\:{\bigg(k \bigg) }^{\dfrac{1}{p} + \dfrac{1}{q} } = {\bigg(k \bigg) }^{ - \dfrac{1}{r} }$

$\rm \implies\:\dfrac{1}{p} + \dfrac{1}{q} = - \dfrac{1}{r}$

$\bf \implies\:\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = 0$

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Additional Information :-

$\boxed{\tt{ \: \: {x}^{p} \: \div \: {x}^{q} \: = \: {x}^{p \: - \: q} \: \: }}$

$\boxed{\tt{ {( {x}^{p} )}^{q} \: = \: {x}^{pq} \: }}$

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

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