Solve for $x$, correct to two significant figures, the equation: $4^{x}-2^{x+1}-3=0$ Solve for $x$, correct to two significant figures, the equation: 
$$4^{x}-2^{x+1}-3=0$$ 
My answer: $x\log4=\log3+(x+1)\log2 \Rightarrow 0.602x-0.301x=0.477+0.301 \Rightarrow x = 2.6$ (Conflicting with book answer)
Answer in book: $x=1.6$

$f(x) = 4^x - 2^{x+1} - 3 = (2^x)^2 - 2^{x+1} - 3 = (2^x)^2 - 2(2^x) - 3$
Let $y = 2^x \Rightarrow f(x) = y^2 - 2y - 3$
$f(x) = 0 \Rightarrow y^2 - 2y - 3 = 0 \Rightarrow  (y-3)(y+1) = 0$
The roots for this equation are $y = 3$ or $y = -1$. Since $y = 2^x$, the solutions in terms of $x$ are $2^x = 3$ or $2^x = -1$ Therefore, the solutions are either $\log _2 \left( 3 \right) $ or $\log _2 \left( -1 \right) $. The latter is undefined, so your answer is $\log _2 \left( 3 \right) $ which is equal to $1.6$ to 2 significant figures. Therefore, $x = 1.6$ and the answer in the book is correct.
