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

The greatest number among $\root 3 \of 9 ,\root 4 \of {11} ,\root 6 \of {17} $ is

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

If $x + \sqrt {({x^2} + 1)} = a,$ then $x =$

${{{{2.3}^{n + 1}} + {{7.3}^{n - 1}}} \over {{3^{n + 2}} - 2{{(1/3)}^{l - n}}}} = $

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