Question

if 2^a=3^b=12^c prove that 1/b+2/a=1/c​

Answer 1

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

$\rm :\longmapsto\: {2}^{a} = {3}^{b} = {12}^{c}$

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

$\rm :\longmapsto\:\dfrac{1}{b} + \dfrac{2}{a} = \dfrac{1}{c}$

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

Given that

$\rm :\longmapsto\: {2}^{a} = {3}^{b} = {12}^{c}$

Let assume that

$\rm :\longmapsto\: {2}^{a} = {3}^{b} = {12}^{c} = x$

$\rm :\longmapsto\: {2}^{a} = x$

We know that

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

So, using this

$\rm :\implies\:2 = {\bigg(x\bigg) }^{\dfrac{1}{a} }$

Also,

$\rm :\longmapsto\: {3}^{b} = x$

$\rm :\implies\:3 = {\bigg(x\bigg) }^{\dfrac{1}{b} }$

Also,

$\rm :\longmapsto\: {12}^{c} = x$

$\rm :\implies\:12 = {\bigg(x\bigg) }^{\dfrac{1}{c} }$

$\rm :\longmapsto\:2 \times 2 \times 3= {\bigg(x\bigg) }^{\dfrac{1}{c} }$

$\rm :\longmapsto\:{\bigg(x\bigg) }^{\dfrac{1}{a} } \times {\bigg(x\bigg) }^{\dfrac{1}{a} } \times {\bigg(x\bigg) }^{\dfrac{1}{b} } = {\bigg(x\bigg) }^{\dfrac{1}{c} }$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:{\bigg(x\bigg) }^{\dfrac{1}{a} + \dfrac{1}{a} + \dfrac{1}{b} } = {\bigg(x\bigg) }^{\dfrac{1}{c} }$

$\rm :\longmapsto\:{\bigg(x\bigg) }^{\dfrac{2}{a} + \dfrac{1}{b} } = {\bigg(x\bigg) }^{\dfrac{1}{c} }$

We know,

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

So,

$\rm :\longmapsto\:\dfrac{1}{b} + \dfrac{2}{a} = \dfrac{1}{c}$

Hence, Proved

Additional Information :-

$\boxed{ \rm{ {a}^{x} \div {a}^{y} = {a}^{x - y}}}$

$\boxed{ \rm{ {a}^{0} = 1}}$

$\boxed{ \rm{ {a}^{ - x} = \frac{1}{ {a}^{x} } }}$

$\boxed{ \rm{ {a}^{x} \times {b}^{x} = {(ab)}^{x}}}$

$\boxed{ \rm{ {a}^{ - 1} = \frac{1}{a}}}$