How do you solve #2^(x^2+x) - 4^(1+x) = 0#?

1 Answer
George C.
Aug 9, 2015

Express both terms as powers of #2# and rearrange to get a quadratic in #x#, giving solutions: #x = -1# or #x = 2#

Explanation:

Add #4^(1+x)# to both sides to get:

#2^(x^2+x) = 4^(1+x) = (2^2)^(1+x) = 2^(2(1+x)) = 2^(2x+2)#

So #x^2+x = 2x+2#

Subtract #2x+2# from both sides to get:

#0 = x^2-x-2 = (x-2)(x+1)#

So #x = -1# or #x = 2#

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