How do you solve #4^ { x ^ { 2} + 4x } = 2^ { - 6}#?

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Answer 1 · 2018-04-10T19:33:10

#color(blue)(x=-3) and color(blue)(x=-1)#

#4^(x^2+4x)=2^(-6)#

#4=2^2#

Substituting this in equation:

#2^(2(x^2+4x))=2^(-6)#

If the bases are equal, then the exponents are equal:

#:.#

#2(x^2+4x)=-6#

Solving for #x#:

#x^2+4x=-3#

#x^2+4x+3=0#

Factor:

#(x+3)(x+1)=0=>color(blue)(x=-3) and color(blue)(x=-1)#