How do you solve #4^ { x - 1} \cdot 8= 2^ { - 4x }#?

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Answer 1 · 2017-08-08T22:34:16

Real solution:

#x = -1/6#

Complex solutions:

#x = -1/6+(npi)/(3 ln 2)i" "# for any integer #n#

Note that:

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

So we find:

#2^(2x+1) = 2^(2x-2)*2^3 = 4^(x-1)*8 = 2^(-4x)#

#color(white)()#
Real solutions

Note that #2^x# is one to one as a real valued function of real numbers, so in order that:

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

we must have:

#2x+1 = -4x#

for real values of #x#.

Hence:

#6x = -1#

So:

#x = -1/6#

#color(white)()#
Complex solutions

Note that:

#e^(2npii) = 1" "# for any integer #n#

Hence we find:

#2^((2npii)/(ln 2)) = (e^(ln 2))^((2npii)/(ln 2)) = e^(2npii) = 1#

Going back to our equation:

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

We can deduce that:

#2x+1 = -4x+(2npii)/ln 2#

Hence:

#6x = -1+(2npii)/ln 2#

So:

#x = -1/6+(npi)/(3 ln 2)i" "# for any integer #n#.