Question: If $x^a = y^b = z^c$ and $y^3 = xz$, show that: $\dfrac{3}{b} = \dfrac{1}{a} + \dfrac{1}{c}$

Solution:

Given: $x^a = y^b = z^c$ and $y^3 = xz$

To prove: $\dfrac{3}{b} = \dfrac{1}{a} + \dfrac{1}{c}$

We have,

$x^a = y^b$

$or, x = y^{\frac{b}{a}}$ - (i)

And,

$z^c = y^b$

$or, z = y^{\frac{b}{c}}$ -(ii)

Also,

$y^3 = xz$

[ Put the values of x and z from equations (i) and (ii) ]

$or, y^3 = y^{\frac{b}{a}} × y^{\frac{b}{c}}$

$or, y^3 = y^{\frac{b}{a} + \frac{b}{c}}$

$or, y^3 = y^{ \frac{bc + ab}{ac}}$

$or, 3 = \dfrac{b(a +c)}{ac}$

$or, \dfrac{3ac}{b} = a + c$

[ Dividing by ac ]

$or, \dfrac{3ac}{abc} = \dfrac{a}{ac} + \dfrac{c}{ac}$

$or, \dfrac{3}{b} = \dfrac{1}{c} + \dfrac{1}{a}$

$\therefore \dfrac{3}{b} = \dfrac{1}{a} + \dfrac{1}{c}$

#proved


Related Notes and Solutions:

Here is the website link to the notes of Indices.

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