જો {a^{1/x}} = {b^{1/y}} = {c^{1/z}} અને {b^2 | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b) ${a^{1/x}} = {b^{1/y}} = {c^{1/z}} = k\,({\rm{say}}) \Rightarrow a = {k^x},\,b = {k^y},\,c = {k^z}$

${b^2} = ac \Rightarrow {({k^y})^2} = {k^x}

Vedclass · Accepted Answer

$2y$

Vedclass · Answer

(b) ${a^{1/x}} = {b^{1/y}} = {c^{1/z}} = k\,({\rm{say}}) \Rightarrow a = {k^x},\,b = {k^y},\,c = {k^z}$

${b^2} = ac \Rightarrow {({k^y})^2} = {k^x}.{k^z}$ $ \Rightarrow $ ${k^{2y}} = {k^{x + z}} \Rightarrow x + z = 2y$.

જો ${a^{1/x}} = {b^{1/y}} = {c^{1/z}}$ અને ${b^2} = ac$ તો $x + z = $

Similar Questions

જો $x \ne 0 $ તો ${\left( {{{{x^l}} \over {{x^m}}}} \right)^{({l^2} + lm + {m^2})}}$${\left( {{{{x^m}} \over {{x^n}}}} \right)^{({m^2} + nm + {n^2})}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{({n^2} + nl + {l^2})}}=$

${{\sqrt 2 } \over {\sqrt {(2 + \sqrt 3 )} - \sqrt {(2 - \sqrt 3 } )}} = $

જો ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{3 \over 2}} \right)^{2 - 2x}},$ તો $x =$

$\root 4 \of {(17 + 12\sqrt 2 )} = $

જો ${a^x} = {(x + y + z)^y},{a^y} = {(x + y + z)^z}$, ${a^z} = {(x + y + z)^x},$ તો