# 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