How do you solve #2^x^2 = 32(2^(4x))#?

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Answer 1 · 2015-10-26T21:27:25

#x= 5 or -1#

Looking at your format I assume you meant to write:

#2^ (x^2) = 32 (2^(4x))#

#=> (2^ (x^2))/(2^(4x)) = 32#

#=> 2^((x^2 - 4x)) = 32#

Taking logs:

#=>(x^2 - 4x)log(2) = log(32)#

#=> x^2 - 4x = (log(32))/(log(2))#

This gives us a quadratic of:

# x^2 - 4x - (log(32))/(log(2)) =0#

But # (log(32))/(log(2)) = 5# giving:

#x^2 - 4x - 5=0#

#(x -5)(x+1)=0#

so #x= 5 or -1#
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Check

#x(x+1)-5(x+1)=x^2+x-5x-5#

#x^2-4x-5#