How do you solve #4^(2x) + 4^x + 12 = 0#?

1 Answer
Jim H
Aug 5, 2015

There is no real solution.

Explanation:

#4^(2x) + 4^x + 12 = 0#

#(4^x)^2 + 4^x + 12 = 0#

You can do a replacement: #u = 4^x#

#u^2 + u +12=0#

#u = (-1+-sqrt(1^2-4(1)(12)))/2#

# = (-1+-sqrt(-47))/2#

#u# is imaginary so we would need #4^x# is imaginary and whie that is certainly possible for imaginary #x#, it is not normally handled in a precalculus study of mathematics.

There is no real solution.

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