How do you solve #(1) ^x = 3^(x-6)#?
1 Answer
May 17, 2016
Real solution:
#x=6#
Complex solutions:
#x = 6 + (2k pi i)/ln 3# for any integer#k in ZZ#
Explanation:
For any value of
So our equation simplifies to:
#3^(x-6) = 1#
If
#3^(x-6) = 3^0 = 1#
satisfying the equation.
As a Real valued function of Reals
Complex solutions
If
#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
#x = 6 + (2k pi i)/ln 3#
for any integer