Question

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

Answer 1

=fales

Answer 2

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

Answer 3

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

Answer 4

im guessing that's \( 4^{x + 1}\)

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

another method:

2^x = 2^(2x +2)

x = 2x + 2
2x - x = -2
x = -2

same thing :D

Answer 11

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