`@` `\text {Ans}`
`\downarrow`
`32^x \div 16^x = 128`
`=> (2^5)^x \div (2^4)^x = 2^7`
`=> 2^(5x) \div 2^(4x) = 2^7`
`=> 2^(5x - 4x) = 2^7`
`=> 2^x = 2^7`
`=> x = 7`
Vậy, `x = 7.`
`32^{x} : 16^{x} = 128`
`=>(2^5)^{x} : (2^{4})^{x} = 2^7`
`=>2^{5x} : 2^{4x} = 2^{7}`
`=>2^{x} = 2^{7}`
`=>x = 7`
Vậy `x=7`