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

1 Answer
Dec 8, 2016

Answer:

#x = -1 + ((2k+1)pi)/(ln(3)) i" "# for any integer #k#

Explanation:

For any Real value of #x# we have #(1/3)^x > 0#. So there is no Real solution.

There are Complex solutions. Consider Euler's identity:

#e^(ipi) = -1#

Hence:

#e^(i(2k+1)pi) = -1" "# for any integer #k#

Now:

#(1/3)^x = (e^(ln (1/3)))^x = e^(-xln(3))#

So if #" "x = -1 + ((2k+1)pi)/(ln(3)) i" "# then we find:

#(1/3)^x = e^(-ln(3)(-1 + ((2k+1)pi)/(ln(3)) i))#

#color(white)((1/3)^x) = e^(ln(3) + (2k+1)pi i)#

#color(white)((1/3)^x) = e^(ln(3)) * e^((2k+1)pi i)#

#color(white)((1/3)^x) = 3 * (-1)#

#color(white)((1/3)^x) = -3#