Question

find 1/x+1/y+1/z? with the given conditions.​

Answer 1

Question :-

11^x = 3^y = 33^z, then (1/x) + (1/y) + (1/z)

Answer :-

Value of (1/x) + (1/y) + (1/z) is 2/z

Explanation :-

11^x = 3^y = 33^z

Let 11^x = 3^y = 33^z = k

Finding the value of 11 in terms of k

$\mathsf{11^x = k}$

$\mathsf{ \implies11 = \sqrt[x]{k} }$

$\mathsf{ \implies11 = k^{ \dfrac{1}{x} } }$

$\boxed{ \bf \because \sqrt[m]{a} = a^{ \dfrac{1}{m} } }$

Finding the value 3 in terms of k

$\mathsf{3^y = k}$

$\mathsf{ \implies 3= \sqrt[y]{k} }$

$\mathsf{ \implies 3 = k^{ \dfrac{1}{y} } }$

$\boxed{ \bf \because \sqrt[m]{a} = a^{ \dfrac{1}{m} } }$

Finding the value of 33 in terms of k

$\mathsf{33^z = k}$

$\mathsf{ \implies 33= \sqrt[z]{k} }$

$\mathsf{ \implies 33 = k^{ \dfrac{1}{z} } }$

$\boxed{ \bf \because \sqrt[m]{a} = a^{ \dfrac{1}{m} } }$

Finding the value of (1/x) + (1/y) + (1/z)

We know that

$\bf{33 = 11 \times 3}$

Substituting the value of 33, 11 and 3 in the above equation

$\mathsf{ \implies k^{ \dfrac{1}{z} } = k^{ \dfrac{1}{x} } \times k^{ \dfrac{1}{y}} }$

$\mathsf{ \implies k^{ \dfrac{1}{z} } = k^{ \dfrac{1}{x} + \dfrac{1}{y} } }$

$\boxed{ \bf \because a^m \times a^n = a^{m+n} }$

$\mathsf{ \implies \dfrac{1}{z} = \dfrac{1}{x} + \dfrac{1}{y} }$

$\boxed{ \bf \because \ if \ a^m = a^n \ then \ m = n }$

$\mathsf{ \implies 0 = \dfrac{1}{x} + \dfrac{1}{y} - \dfrac{1}{z} }$

Adding 2/z on both sides

$\mathsf{ \implies \dfrac{2}{z} = \dfrac{1}{x} + \dfrac{1}{y} - \dfrac{1}{z} + \dfrac{2}{z} }$

$\mathsf{ \implies \dfrac{2}{z} = \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} }$

$\mathsf{ \implies \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{2}{z} }$

∴ the value of (1/x) + (1/y) + (1/z) is 2/z

Answer 2

Given

${11}^{x} = {3}^{y} = {33}^{z}$

To find

$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$

Explanation

let us assume that

$= > {11}^{x} = {3}^{y} = {33}^{z} = t$

Now equating each and every term to t

$= > {11}^{x} = t$

$= > 11 = {t}^{ \dfrac{1}{x} } ........(1)$

$= > {3}^{y } = t$

$= > 3 = {t}^{ \dfrac{1}{y} } .........(2)$

$= > {33}^{z} = t$

$= > 33 = {t}^{ \dfrac{1}{z} } .........(3)$

we know that

3×11=33

Now substitute the values from (1) ,(2),(3)

then

${t}^{ \dfrac{1}{y} } \times {t}^{ \dfrac{1}{x} } = {t}^{ \dfrac{1}{z} }$

$\underline{\sf a^m×a^n=a^m+n}$

$= > {t}^{ \dfrac{1}{x} + \dfrac{1}{y} } = {t}^{ \dfrac{1}{z} }$

Now as the bases are equal powers can be equated

$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z} ..........(4)$

now consider

$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$

from eq (4)

$= > \dfrac{1}{z} + \dfrac{1}{z}$

$= > \dfrac{1 + 1}{z} = \dfrac{2}{z}$

$\boxed {\sf \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{2}{z}}$