How do you solve 12(1-4^x)=1812(1−4x)=18?

1 Answer
George C.
Jul 31, 2016

x=-1/2+((2k+1)pi i)/ln(4)x=−12+(2k+1)πiln(4) for any integer kk

Explanation:

Divide both sides by 1212 to get:

1-4^x = 18/12 = 3/21−4x=1812=32

Add 4^x-3/24x−32 to both sides to get:

-1/2 = 4^x−12=4x

For any Real value of xx, 4^x > 04x>0, so it cannot equal -1/2−12.

If we had wanted 4^x = 1/24x=12 then we could have found:

2^(-1) = 1/2 = 4^x = (2^2)^x = 2^(2x)2−1=12=4x=(22)x=22x

and hence solution x=-1/2x=−12.

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

Let x=-1/2+((2k+1)pi i)/ln(4)x=−12+(2k+1)πiln(4) for any integer kk

Then:

4^x = 4^(-1/2+((2k+1)pi i)/ln(4))4x=4−12+(2k+1)πiln(4)

=4^(-1/2)*4^(((2k+1)pi i)/ln(4))=4−12⋅4(2k+1)πiln(4)

=1/2*(e^ln(4))^(((2k+1)pi i)/ln(4))=12⋅(eln(4))(2k+1)πiln(4)

=1/2*e^(ln(4)*((2k+1)pi i)/ln(4))=12⋅eln(4)⋅(2k+1)πiln(4)

=1/2*e^((2k+1)pi i)=12⋅e(2k+1)πi

=1/2*(e^(pi i))^(2k+1)=12⋅(eπi)2k+1

=1/2*(-1)^(2k+1)=12⋅(−1)2k+1

=-1/2=−12

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