Question 657728
4^y+4^(2*y+1)=1

Simplify the question by making a substitution.

Let x = 4^y
{{{4^y + 4^(2y+1)=1}}}
{{{4^y + 4(4^y)^2=1}}}
{{{4x^2 + x = 1}}}
{{{4x^2 + x - 1 = 0}}}

*[invoke quadratic "x", 4, 1, -1 ]

0.390388203202208, -0.640388203202208

sub into x = 4^y
logx = ylog4
y = logx/log4
y1 = -0.6785
y2 = invalid (can't log negative numbers)

The answer is approximately -0.6785.  If you want exact formula you'll need to use the quadratic formula to solve and leave in exact numbers.