How do you solve #12(1-4^x)=18#?

1 Answer
George C.
Jul 31, 2016

#x=-1/2+((2k+1)pi i)/ln(4)# for any integer #k#

Explanation:

Divide both sides by #12# to get:

#1-4^x = 18/12 = 3/2#

Add #4^x-3/2# to both sides to get:

#-1/2 = 4^x#

For any Real value of #x#, #4^x > 0#, so it cannot equal #-1/2#.

If we had wanted #4^x = 1/2# then we could have found:

#2^(-1) = 1/2 = 4^x = (2^2)^x = 2^(2x)#

and hence solution #x=-1/2#.

We can make this into a set of Complex solutions for our original problem by adding an odd multiple of #(pi i)/ln(4)#

Let #x=-1/2+((2k+1)pi i)/ln(4)# for any integer #k#

Then:

#4^x = 4^(-1/2+((2k+1)pi i)/ln(4))#

#=4^(-1/2)*4^(((2k+1)pi i)/ln(4))#

#=1/2*(e^ln(4))^(((2k+1)pi i)/ln(4))#

#=1/2*e^(ln(4)*((2k+1)pi i)/ln(4))#

#=1/2*e^((2k+1)pi i)#

#=1/2*(e^(pi i))^(2k+1)#

#=1/2*(-1)^(2k+1)#

#=-1/2#

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