If {a^{x - 1}} = bc,{b^{y - 1}} = ca,{c^{z - | vedclass

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

Question

(a) ${a^{x - 1}} = bc \Rightarrow {a^x} = abc$

$	herefore {a^x} = {b^y} = {c^z} = abc = k\,({m{say}})$

$ \Rightarrow $$a = {k^{1/x}} \Rightar

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(a) ${a^{x - 1}} = bc \Rightarrow {a^x} = abc$

$	herefore {a^x} = {b^y} = {c^z} = abc = k\,({m{say}})$

$ \Rightarrow $$a = {k^{1/x}} \Rightarrow {1 \over x} = {\log _k}a$; $\sum\limits_{}^{} {{1 \over x} = {{\log }_k}a + {{\log }_k}b + {{\log }_k}c} = {\log _k}abc = {\log _{abc}}abc = 1$.

If ${a^{x - 1}} = bc,{b^{y - 1}} = ca,{c^{z - 1}} = ab,$then $\sum {(1/x) = } $

Similar Questions

The rationalising factor of ${a^{1/3}} + {a^{ - 1/3}}$ is

If ${a^x} = {b^y} = {(ab)^{xy}},$ then $x + y = $

Solution of the equation ${9^x} - {2^{x + {1 \over 2}}} = {2^{x + {3 \over 2}}} - {3^{2x - 1}}$

Solution of the equation ${4.9^{x - 1}} = 3\sqrt {({2^{2x + 1}})} $ has the solution

Solution of the equation $\sqrt {(x + 10)} + \sqrt {(x - 2)} = 6$ are