Question

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Answer 1

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

Given :-

$\rm :\longmapsto\: {2}^{x} = {5}^{y} = {10}^{z}$

To Prove :-

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

Identity Used :-

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

$\rm :\longmapsto\: {a}^{m} \times {a}^{n} = {a}^{m + n}$

$\rm :\longmapsto\:If \: {x}^{m} = {x}^{n} \: \implies \: m \: = \: n$

Calculations :-

Given that

$\rm :\longmapsto\: {2}^{x} = {5}^{y} = {10}^{z}$

Let assume that

$\rm :\longmapsto\: {2}^{x} = {5}^{y} = {10}^{z}= k$

So,

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

$\bf\implies \:2 = {\bigg(k\bigg) }^{\dfrac{1}{x} } - - - (1)$

Also,

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

$\bf\implies \:5 = {\bigg(k\bigg) }^{\dfrac{1}{y} } - - - (2)$

Also,

$\rm :\longmapsto\: {10}^{z} = k$

$\bf\implies \:10 = {\bigg(k\bigg) }^{\dfrac{1}{z} }$

$\bf\implies \:2 \times 5 = {\bigg(k\bigg) }^{\dfrac{1}{z} }$

On substituting the values of 2 and 5, we get

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

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

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

Hence, Proved

Additional Information :-

$\rm :\longmapsto\: {x}^{0} = 1$

$\rm :\longmapsto\: {x}^{ - n} = \dfrac{1}{ {x}^{n} }$

$\rm :\longmapsto\: {a}^{m} \div {a}^{n} = {a}^{m - n}$

$\rm :\longmapsto\: {( {a}^{m} )}^{n} = {a}^{mn}$

Answer 2

2^x=5^ʏ=10^ᴢ

1/x+1/ʏ=1/ᴢ

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