proove the following.........
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proove the following.........
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Answer -
» Given :
[tex]x^p= y^q = z^r\\ \\ \\ \frac{y}{x} = \frac{z}{y} [/tex]
» To prove:
[tex]\frac{2}{q} = \frac{1}{p} +\frac{1}{r} [/tex]
» Solution:
assume ,
[tex]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} }[/tex]
Given,
[tex]\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[/tex]