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`\green{((2^x + 22^x + 222^x))/((3^x + 33^x + 333^x)) =( 9)/(4)}`​

Answer 1

Answer:-

`\bold{\blue{(2^(x)+(22)^(x)+(222)^(x))/(3^(x)+(33)^(x)+(333)^(x))=(9)/(4)}  }`

`\bold{\pink{(2^(x)+(22)^(x)+(222)^(x))/(3^(x)+(33)^(x)+(333)^(x))=[(4)/(9)]^(-1) }}`

`\bold{\green{(2^(x)(1+11^(x)+111^(x)))/(3^(x)(1+11^(x)+111^(x)))=(((2)/(3))^{{{2}}})^{{{-1}}}   }}`

`\bold{\orange{\frac{2^(x)\cancel{(1+11^(x)+111^(x))}}{3^(x)\cancel{(1+11^(x)+111^(x))}}=((2)/(3))^(-2)   }}`

`\bold{\purple{(2^(x))/(3^(x))=((2)/(3))^(-2)   }}`

`\bold{\red{((2)/(3))^(x) =((2)/(3))^(-2)   }}`

`\bold{\green{ x=-2 } }`

Answer 2

Explanation:

simplify upper n lower fraction of LHS:

2^x+22^x+222^x

= 2^x*1 + (2*11)^x + (2*111)^x

= 2^x*1 + 2^x*11^x + 2^x*111^x

= 2^x*(1+11^x+111^x)

3^x+33^x+333^x

= 3^x*1 + (3*11)^x + (3*111)^x

= 3^x*1 + 3^x*11^x + 3^x*111^x

= 3^x*(1+11^x+111^x)

so LHS

= 2^x*(1+11^x+111^x) / (3^x*(1+11^x+111^x))

= 2^x / 3^x

= (2/3)^x

RHS = 9/4

= (3^2/ 2^2)

= (3/2)^2

= (2/3)^(-2)

so x = -2