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प्रश्न
`2^2x-3.2^((x+2))+32=0`
उत्तर
`2^2x-3.2^((x+2))+32=0`
⇒ `(2^x)^2-3.2^x.2^2+32=0`
Let `2^x` be y.
∴`y^2-12y+32=0`
⇒`y^2-8y-4y+32=0`
⇒ `y(y-8)-4(y-8)=0`
⇒`(y-8)=01 or (y-4)=0`
⇒`y=8 or y=4`
∴`2^x=8 or 2^x=4`
⇒ `2^x=2^3 or 2^x=2^2`
⇒`x-2 or 3`
Hence, 2 and 3 are the roots of the given equation.
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