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

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Answer 1 · 2016-06-27T19:45:26

#x=1/2#.

You have that #16=4^2#, you can write

#16^(x+1)=4^(2(x+1))# then we have

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

this equations is verified when the exponents are equal

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

#2x+2=4x+1#

#2x=1#

#x=1/2#.

We can verify if this solution is correct substituting in the original equation

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

#16^(3/2)=4^3#

#(sqrt(16))^3=4^3#

here we have to be careful because the square root of #16# can be #\pm4# but only the solution with the #+# is valid.

#4^3=4^3#.