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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