How do you solve #(1) ^x = 3^(x-6)#?

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Answer 1 · 2016-05-17T21:59:14

Real solution:

#x=6#

Complex solutions:

#x = 6 + (2k pi i)/ln 3# for any integer #k in ZZ#

For any value of #x#, we have #1^x = 1#

So our equation simplifies to:

#3^(x-6) = 1#

If #x = 6# then:

#3^(x-6) = 3^0 = 1#

satisfying the equation.

As a Real valued function of Reals #f(t) = 3^t# is one to one, so this is the unique Real solution.

Complex solutions

If #k# is any integer then:

#1 = e^(2k pi i) = (e^(ln 3))^((2kpi i)/(ln 3)) = 3^((2kpi i)/(ln 3))#

So:

#3^(t+(2k pi i)/ln 3) = 3^t * 3^((2k pi i)/ln 3) = 3^t#

Hence the equation #(1)^x = 3^(x-6)# has solutions:

#x = 6 + (2k pi i)/ln 3#

for any integer #k in ZZ#