SOLUTION: how to get the solution of this equation. 2^2x+2^x-1=0

Question 584741: how to get the solution of this equation. 2^2x+2^x-1=0
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
2^2x is the same as (2^x)^2
your equation of:
2^2x + 2^x - 1 = 0 becomes:
(2^x)^2 + (2^x) - 1 = 0
unfortunately, this doesn't factor.
it does have real roots though, so you can factor it using the quadratic formula.
using the quadratic formula, i get:
2^x = -1.6180339887
or:
2^x = .6180339887
to solve for x, we take the log of both sides of each equation to get:
log(2^x) = log(-1.618033989)
by the law of logarithms that states that log(x^a) = a*log(x), we get:
x*log(2) = log(-1.618033989)
unfortunately, we can't get the log of a negative number, so this solution is invalid.
we still have our other solution to test though.
that one is:
2^x = .6180339887
we take the log of both sides of this equation to get:
log(2^x) = log(.6180339887)
by the laws of logarithms again, we get:
x*log(2) = log(.6180339887)
we divide both sides of this equation by log(2) to get:
x = log(.6180339887) / log(2) which we can solve to get:
x = -.6942419136
we substitute that value of x in our original equation of:
2^(2x) + 2^x - 1 = 0 to get:
2^(2*(-.6942419136)) + 2^(-.6942419136) - 1 = 0
solve using your calculator to get:
0 = 0
this confirms the value of -.6942419136
the key to this problem was that 2^2x is the same as (2^x)^2.
our variable becomes (2^x) and we solve just like we solve any other quadratic equation except we then have to solve to find the exponent of x.
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