solve and check 2^x = 4^x + 1

Question

solve and check  2^x = 4^x + 1

anonymous · Answer

=fales

anonymous · Answer

$$
x=-\frac{i \pi }{3 \ln
   (2)},\quad x=\frac{i \pi }{3 \ln
   (2)}
$$

anonymous · Answer

The options are -1, -2, 2, or 0.

lgbasallote · Answer

im  guessing that's $ 4^{x + 1}$

lgbasallote · Answer

so it would be $\Large  2^x = 4^{x+1} \rightarrow 2^x = 2^{2(x+1)}$ any ideas now?

anonymous · Answer

None of -1, -2, 2, or 0. is a solution if your equation is
$$
2^x = 4^x +1
$$

australopithecus · Answer

oh oh it is

2^(x) = 4^(x+1)

xln(2) = (x+1)ln(4)
xln(2) = xln(4) + ln(4)
xln(2) - xln(4) = ln(4)
x(ln(2) - ln(4)) = ln(4)
x = ln(4)/(ln(2) - ln(4)
x = -2

australopithecus · Answer

if you need to me to explain my method I can :)

australopithecus · Answer

@lgbasallote thanks for pointing out the correct form of the question

lgbasallote · Answer

another method:

2^x = 2^(2x +2)

x = 2x + 2
2x - x = -2
x = -2

same thing :D

anonymous · Answer

Please put parentheses where they belong. There is a big difference between
4^x+1 and 4^(x+1)