How do you solve #64= 32^ { - x + 4}#?

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Answer 1 · 2017-01-03T00:02:55

Real solution:

#x = 14/5#

Complex solutions:

#x = 14/5+(2kpii)/(5 ln(2))" "# for any integer #k#

Note that #64=2^6# and #32=2^5#, so we have:

#2^6 = 64 = 32^(-x+4) = (2^5)^(-x+4) = 2^(5(-x+4))#

As a real valued function of real numbers, #x -> 2^x# is one to one.

So (for Real solutions) we must have:

#6 = 5(-x+4) = -5x+20#

Add #5x-6# to both ends to get:

#5x = 14#

Divide both sides by #5# to get:

#x = 14/5#

Further note that:

#e^(2kpii) = 1" "# for any integer #k#

So we find:

#2^((2kpii)/(ln2)) = (e^(ln(2)))^((2kpii)/(ln(2))) = e^(ln(2)*((2kpii)/(ln(2)))) = e^(2kpii) = 1#

Hence we have additional Complex solutions corresponding to:

#6 = 5(-x+4)+(2kpii)/(ln(2))#

Hence:

#5x = 14+(2kpii)/(ln(2))#

Hence:

#x = 14/5+(2kpii)/(5 ln(2))" "# for any integer #k#