$\displaystyle 2^x+28*(2^{-x})=11 $ show how this can be rewritten as $\displaystyle (2^x)^2-11*2^x+28=0$ thanks
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Originally Posted by BabyMilo $\displaystyle 2^x+28*(2^{-x})=11 $ show how this can be rewritten as $\displaystyle (2^x)^2-11*2^x+28=0$ thanks $\displaystyle 2^x+28*(2^{-x})=11~\implies~2^x+\dfrac{28}{2^x}-11=0 $ Now multiply both sides of the equation by $\displaystyle 2^x$ and you'll get the given result at once.
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