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: $x=1.6$
 A: You made a mistake in your first step. Take a look at the following
$$\eqalign{
  & {4^x} - {2^{x + 1}} - 3 = 0  \cr 
  & {\left( {{2^x}} \right)^2} - 2\left( {{2^x}} \right) - 3 = 0  \cr 
  & \left( {{2^x} - 3} \right)\left( {{2^x} + 1} \right) = 0  \cr 
  & \left\{ \matrix{
  {2^x} = 3 \hfill \cr 
  {2^x} =  - 1,\,\,\,\text{impossible} \hfill \cr}  \right.  \cr 
  & x\ln 2 = \ln 3  \cr 
  & x = {{\ln 3} \over {\ln 2}} = \log_{2}3 =1.584  \cr} $$
A: You made a mistake:
$$\log(a+b)\neq \log(a)+\log(b)$$
You are probably confusing this with the correct law:
$$\log(ab)=\log(a)+\log(b)$$
The way to solve this problem is to let $u=2^x$, then the equation becomes:
$$u^2-2u-3=0$$
Solve this for $u$, then take logs to get $x$
A: $\ln(x+y)\neq \ln(x)+\ln(y)$, as you have done.
Instead, make the substitution $y=2^x$. Then you will get a quadratic equation
$y^2-2y-3=0$ (Do you see why?)
By solving for $y$, you can solve for $x$ using logarithms.
A: $f(x) = 4^x - 2^{x+1} - 3 = (2^x)^2 - 2^{x+1} - 3 = (2^x)^2 - 2(2^x) - 3$
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.
A: Put 2^x = Z and the equation reduces to 
Z^2-2Z-3=0
Thus, Z=3 or -1
The only solution is 2^x=3, which means x=Log2(3)=1.584
