Question

plz solve it plz the attached file​

Answer 1

QUESTION:

• If x, y and y are positive real numbers and p, q and r are natural numbers such that x^p = y^q = z^r and y/x = z/y, then prove that 2/q = 1/p + 1/r

ANSWER:

Given:

• x, y and y are positive real numbers
• p, q and r are natural numbers
• x^p = y^q = z^r
• y/x = z/y

To Prove:

• 2/q = 1/p + 1/r

Proof:

We are given that,

$\implies\sf x^p=y^q=z^r$

Let these be equal to a constant 'k'.

That means,

$\implies\sf x^p=y^q=z^r=k$

So,

$\implies\sf x^p=k$

$\implies\sf x=k^{\frac{1}{p}}- - - -(1)$

Similarly,

$\implies\sf y^q=k$

$\implies\sf y=k^{\frac{1}{q}}- - - -(2)$

And,

$\implies\sf z^r=k$

$\implies\sf z=k^{\frac{1}{r}}- - - -(3)$

Now, we are also given that,

$\implies\sf\dfrac{y}{x}=\dfrac{z}{y}$

On cross-multiplying,

$\implies\sf y \times y=x \times z$

So,

$\implies\sf y^2=xz$

Now, substituting values of x, y and z from (1), (2) & (3),

$\implies\sf (k^{\frac{1}{q}})^2=k^{\frac{1}{p}}\times k^{\frac{1}{r}}$

$\implies\sf k^{\frac{2}{q}}=k^{\frac{1}{p}}\times k^{\frac{1}{r}}$

As, a^m × a^n = a^(m+n),

So,

$\implies\sf k^{\frac{2}{q}}=k^{\frac{1}{p}+\frac{1}{r}}$

As, the bases are same we compare the powers,

$\implies\bf \dfrac{2}{q}=\dfrac{1}{p}+\dfrac{1}{r}$

That is what we were to prove.

Hence Proved!!