How do you solve #2^(2x)times4^(4x+8)=64#?

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Answer 1 · 2016-09-18T13:03:20

#x = - 1#

We have: #2^(2 x) times 4^(4 x + 8) = 64#

Let's express the numbers in terms of #2#:

#=> 2^(2 x) times (2^2)^(4 x + 8) = 2^(6)#

Using the laws of exponents:

#=> 2^(2 x) times 2^(8 x + 16) = 2^(6)#

#=> 2^(2 x + 8 x + 16) = 2^(6)#

#=> 2^(10 x + 16) = 2^(6)#

#=> 10 x + 16 = 6#

#=> 10 x = - 10#

#=> x = - 1#

Therefore, the solution to the equation is #x = - 1#.