How do you solve the exponential equation #16^(x+1)=4^(4x+1)#?

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Answer 1 · 2017-03-07T15:51:11

#x=1/2#

the key to these sorts of questions is to change the base numbers so that they are the same.

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

now #16=4^2#

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

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

#=>2x+2=4x+1#

#=>2-1=4x-2x#
#1=2x#

#:.x=1/2#