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

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

Question

(b) ${a^x} = {b^y} = {(ab)^{xy}}$

$ \Rightarrow $$x\ln a = y\ln b = xy\ln (ab) = k\,{\rm{(say)}}$

$\ln a = {k \over x},\,\,\ln b = {k \over y}$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(b) ${a^x} = {b^y} = {(ab)^{xy}}$

$ \Rightarrow $$x\ln a = y\ln b = xy\ln (ab) = k\,{m{(say)}}$

$\ln a = {k \over x},\,\,\ln b = {k \over y}$

$\ln (a\,b) = {k \over {xy}}$$ \Rightarrow $ $\ln a + \ln b = {k \over {xy}}$

$ \Rightarrow$ ${k \over x} + {k \over y} = {k \over {xy}}$

$ \Rightarrow $ ${1 \over x} + {1 \over y} = {1 \over {xy}}$ $ \Rightarrow $ ${{x + y} \over {xy}} = {1 \over {xy}}$;

$	herefore x + y = 1$.

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

Similar Questions

${{{{[4 + \sqrt {(15)} ]}^{3/2}} + {{[4 - \sqrt {(15)} ]}^{3/2}}} \over {{{[6 + \sqrt {(35)} ]}^{3/2}} - {{[6 - \sqrt {(35)} ]}^{3/2}}}} = $

If $x = {2^{1/3}} - {2^{ - 1/3}},$ then $2{x^3} + 6x = $

$\sqrt {(3 + \sqrt 5 )} $ is equal to

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

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