Question

hey guys please solve this problem and don't spam

Answer 1

hope it help u dear......

Answer 2

$\sf{Given : 2^x^+^y = 6^y}\\\\\sf{\implies 2^x^+^y = (2\times 3)^y}\\\\\sf{We\;know\;that : (ab)^n = a^n.b^n}\\\\\sf{\implies 2^x^+^y = (2)^y.(3)^y}\\\\\sf{We\;know\;that : (a)^p^+^q = a^p.a^q}\\\\\sf{\implies (2)^x.(2)^y = (2)^y.(3)^y}\\\\\sf{\implies (2)^x = (3)^y\;-------\;[1]}\\\\Given : 3^x^-^1 = 2^y^+^1\\\\\sf{\implies (3)^x.(3)^-^1 = (2)^y.(2)^1}\\\\\sf{\implies \dfrac{(3)^x}{3} = (2)^y.(2)}\\\\\sf{\implies (3)^x = (2)^y.(6)\;-------\;[2]}$

$\textsf{Multiplying both Equations [1] and [2], We get :}\\\\\sf{\implies (2)^x.(3)^x = (2)^y.(3)^y.(6)}\\\\\sf{\implies (6)^x = (6)^y.(6)}\\\\\sf{\implies (6)^x = (6)^y^+^1}\\\\\sf{We\;know\;that : If\;Bases\;are\;Equal \implies Exponents\;should\;be\;Equal}\\\\\sf{\implies x = y + 1}\\\\Given : \dfrac{log(3) - log(2)}{x - y}\\\\\\\sf{\implies \dfrac{log(3) - log(2)}{y + 1 - y}}\\\\\\\sf{\implies log(3) - log(2)}\\\\\sf{We\;know\;that : log(a) - log(b) = log\big(\dfrac{a}{b}\big)}$

$\sf{\implies log(3) - log(2) = log\big(\dfrac{3}{2}\big)}$