Simply solve for x ?

4ˣ - 2ˣ - 1 = 0

........2007-06-22T02:57:12Z

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4ˣ - 2ˣ - 1 = 0

(2ˣ)² - 2ˣ - 1 = 0

2ˣ = {1 ± √(1 + 4) } / 2

2ˣ = (1±√5)/2

(1-√5)/2 is negative, so use (1+√5)/2

ln(2ˣ) = ln{(1+√5)/2}
x*ln(2) = ln{(1+√5)/2}
x = ln{(1+√5)/2} / ln(2)
= { ln(1+√5) - ln(2) } / ln(2)
= ln(1+√5) / ln(2) - 1

x=.694241913630617301738790266898

I checked the answer, and it works..

Anonymous2007-06-22T03:48:58Z

4ˣ - 2ˣ - 1 = 0 ...... the initial equation
(2²)ˣ - 2ˣ - 1 = 0 ...... attempting to change to a consistent base
2²ˣ - 2ˣ - 1 = 0 ..... simplified
(2ˣ)² - 2ˣ - 1 = 0 .... changed to same base, and into quadratic form
(2ˣ)² - 2ˣ = 1 .... moved the constant to one side
(2ˣ)² - 2ˣ + ¼ = 1 + ¼ .... complete the square
(2ˣ - ½)² = 5/4 .... turn into binomial squared
2ˣ - ½ = ±√(5/4) ..... square root both sides
2ˣ - ½ = ±√5 / 2 ..... denominator in radical
2ˣ = ½ ± √5 / 2 ..... isolate x
2ˣ = (1 ± √5) / 2 .... common denominator
log 2ˣ = log [(1 ± √5) / 2] ... isolate x by logarithms
x log 2 = log [(1 ± √5) / 2] .... exponent in logarithm rule
x = log [(1 ± √5) / 2] / log 2 .... division to isolate x

This is as accurate and all encompassing a solution as you can get.

Working out specific values...
x = 0.6942419... (real)
x = -0.6942419... + i 4.5323601... (complex)

Tubby2007-06-22T02:45:05Z

4ˣ - 2ˣ - 1 = 0
2^2x - 2^x - 2^0 = 0
:> 2x - x - 0 = 0
x = 0

rui_min2007-06-22T02:46:37Z

maybe x is equal to 1/6.

a_ebnlhaitham2007-06-22T02:44:46Z

4ˣ - 2ˣ - 1 = 2ˣ - 1 = 0
2ˣ = 1
ˣ = 0.5