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Solve 4^x + 2^x + 1 = 111.

 Dec 4, 2020
 #1
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 so first we can simplify $4^x$ into $2^2x$. So $2^2x$+$2^x$+$1$=$111$, next is $2^2x$+$2^x$=$110$, add so that it is $2^3x=110, take log on both sides and you get 2.260453

 

pls check if I am correct and don't submit answer until verify

 Dec 4, 2020
 #2
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Let \(u = 2^x\), then we have \(u^2+u+1=111\), or \(u^2 + u - 110 = 0 \) or \((u - 10)(u + 11) = 0\)

 

Assuming ony a real number solution for x means we want u = 10 and \(x=\frac{\ln(10)}{\ln(2)}\) 

 Dec 5, 2020

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