Question

(2^x + 22^x + 222^x)/(3^x + 33^x + 333^x) = 9/4

Answer 1

$\huge{\underline{\underbrace{\mathcal\color{gold}{Answer}}}}$

$\implies$$\bold{\dfrac{2^{x}+(22)^{x}+(222)^{x}}{3^{x}+(33)^{x}+(333)^{x}}=\dfrac{9}{4}   }$ 

$\dashrightarrow$ $\bold{\dfrac{2^{x}+(22)^{x}+(222)^{x}}{3^{x}+(33)^{x}+(333)^{x}}=[\dfrac{4}{9}]^{-1} }$ 

$[/tex] [tex]\bold{\dfrac{2^{x}(1+11^{x}+111^{x})}{3^{x}(1+11^{x}+111^{x})}=((\dfrac{2}{3})^{{{2}}})^{{{-1}}}   }$ 

$\longrightarrow$$\bold{\dfrac{2^{x}\cancel{(1+11^{x}+111^{x})}}{3^{x}\cancel{(1+11^{x}+111^{x})}}=(\dfrac{2}{3})^{-2}   }$ 

$\rightsquigarrow$ $\bold{\dfrac{2^{x}}{3^{x}}=(\dfrac{2}{3})^{-2}   }$ 

$\rightsquigarrow$ $\bold{(\dfrac{2}{3})^{x} =(\dfrac{2}{3})^{-2}   }$ 

$\rightsquigarrow$ $\bold{ x=-2  }$

Answer 2

Step-by-step explanation:

$\tt{\dfrac{2^{x}+(22)^{x}+(222)^{x}}{3^{x}+(33)^{x}+(333)^{x}}=\dfrac{9}{4}   }$ 

$\tt{\dfrac{2^{x}+(22)^{x}+(222)^{x}}{3^{x}+(33)^{x}+(333)^{x}}=[\dfrac{4}{9}]^{-1} }$ 

$\tt{\dfrac{2^{x}(1+11^{x}+111^{x})}{3^{x}(1+11^{x}+111^{x})}=((\dfrac{2}{3})^{{{2}}})^{{{-1}}}   }$ 

$\tt{\dfrac{2^{x}\cancel{(1+11^{x}+111^{x})}}{3^{x}\cancel{(1+11^{x}+111^{x})}}=(\dfrac{2}{3})^{-2}   }$ 

$\tt{\dfrac{2^{x}}{3^{x}}=(\dfrac{2}{3})^{-2}   }$ 

$\tt{(\dfrac{2}{3})^{x} =(\dfrac{2}{3})^{-2}   }$ 

$\tt{ x=-2  }$