How do you solve #8= 16^ { 2x + 1}#?

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Answer 1 · 2017-04-14T06:12:39

#x=-1/8#

#8=16^(2x+1)# can be written as

#2^3=(2^4)^(2x+1)#

or #2^3=2^(4(2x+1))#

or #2^3=2^((8x+4))#

or #8x+4=3#

or #8x=-1#

or #x=-1/8#