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: $x=1.6$

You made a mistake in your first step. Take a look at the following
$$\eqalign{
  & {4^x} - {2^{x + 1}} - 3 = 0  \cr 
  & {\left( {{2^x}} \right)^2} - 2\left( {{2^x}} \right) - 3 = 0  \cr 
  & \left( {{2^x} - 3} \right)\left( {{2^x} + 1} \right) = 0  \cr 
  & \left\{ \matrix{
  {2^x} = 3 \hfill \cr 
  {2^x} =  - 1,\,\,\,\text{impossible} \hfill \cr}  \right.  \cr 
  & x\ln 2 = \ln 3  \cr 
  & x = {{\ln 3} \over {\ln 2}} = \log_{2}3 =1.584  \cr} $$
