How do you solve 12(1−4x)=18?

1 Answer
George C.
Jul 31, 2016

x=−12+(2k+1)πiln(4) for any integer k

Explanation:

Divide both sides by 12 to get:

1−4x=1812=32

Add 4x−32 to both sides to get:

−12=4x

For any Real value of x, 4x>0, so it cannot equal −12.

If we had wanted 4x=12 then we could have found:

2−1=12=4x=(22)x=22x

and hence solution x=−12.

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

Let x=−12+(2k+1)πiln(4) for any integer k

Then:

4x=4−12+(2k+1)πiln(4)

=4−12⋅4(2k+1)πiln(4)

=12⋅(eln(4))(2k+1)πiln(4)

=12⋅eln(4)⋅(2k+1)πiln(4)

=12⋅e(2k+1)πi

=12⋅(eπi)2k+1

=12⋅(−1)2k+1

=−12

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