Question

proove the following.........​

Answer 1

Answer -

» Given :

$x^p= y^q = z^r\\ \\ \\ \frac{y}{x} = \frac{z}{y}$

» To prove:

$\frac{2}{q} = \frac{1}{p} +\frac{1}{r}$

» Solution:

assume ,

$x^p= y^q = z^r = k \\ \\ \\ x^p = k \\ \\ \\ x = k^{\frac{1}{p} } (by\: multiplying\: 1/p \:on\:both\: sides)\\ \\ \\ Similarly, y = k^{\frac{1}{q} } \\ \\ \\ z = k^{\frac{1}{r} }$

Given,

$\frac{y}{x} = \frac{z}{y}\\ \\ \\by\:cross\: multiplication\\ \\ \\ y^2 = xz \\ \\ \\ by \: putting\:the\:values\\ \\ \\ (k^{\frac{1}{q}})^2 = k^{\frac{1}{p}} \times k^{\frac{1}{r}}\\ \\ \\ {\pink{\boxed{law -> (a^m)^n = a^{mn}}}}\\ \\ \\ {\pink{\boxed{law -> a^m \times a^n = a^{m+n}}}}\\ \\ \\ = k^{\frac{2}{q}} = k^{\frac{1}{p} + \frac{1}{r}}\\ \\ \\ we\:know\:that, powers \:are \:equal\:when\:base\:is\:equal\\ \\ \\ \therefore \frac{2}{q} = \frac{1}{p} + \frac{1}{r} \\ \\ \\ Hence \:proved$