fractional exponents and find extraneous solutions in W|A?

Questions
,
1

I am using fractional exponents and I believe there are two answers, but others say (including calculator here in WA) that I am wrong. Please help me to understand. Thank you.

(6x-5)^(4/3)=625
solution creates x= 65/3 ---- everyone agrees this works.
but x= -20?
I say that if you plug it in, you get:
6(-20) - 5 = -125
now take that to the 4th and get 244,140,625.
next, take the 3rd root and get 625.

Now, I realize that if you do not take the 4th first, you get a negative number. BUT as a fractional exponent it would be the third root of the number to the fourth, thus making the inside a positive number, as any real number to an even root would be.

I would love a clear explanation and even better, a reliable, known textbook reference.
Thank you!

2

Whenever you have a fractional power, there is the possibility of extraneous complex roots.

(6x-5)^(4/3) = 625 = 625 * 1

Taking the 3/4 th power and noting that 1^(3/4) = Exp[(3/2)n Pi I], n = 0, 1, 2 … you get

6x-5 = 125 Exp[(3/2)n Pi I]

In[1]:= Solve[6 x - 5 == 125 Exp[3 n Pi I/2], x]

Out[1]= {{x -> 5/6 (1 + 25 E^((3 I n \[Pi])/2))}}

In[2]:= 5/6 (1 + 25 E^((3 I n \[Pi])/2)) /. n -> 0

Out[2]= 65/3

In[3]:= 5/6 (1 + 25 E^((3 I n \[Pi])/2)) /. n -> 1

Out[3]= 5/6 - (125 I)/6

In[4]:= 5/6 (1 + 25 E^((3 I n \[Pi])/2)) /. n -> 2

Out[4]= -20
3

If you want the negative solution to show up you have to write the equation with Surd:

Solve[Surd[6 x - 5, 3]^4 == 625, x, Reals]

or with CubeRoot

Solve[CubeRoot[6 x - 5]^4 == 625, x, Reals]

This is because Mathematica powers are otherwise meant in the complex sense, as the principal root: a^b=Exp[b*Log[a]]. When the base a is negative, the Log is nonreal. Compare

Plot[x^(1/3), {x, -1, 1}]
Plot[CubeRoot[x], {x, -1, 1}]

I always warn my students that the familiar algebraic rules break down when dealing with powers with a negative base and noninteger exponent.

4

Also, compare

Solve[((6 x - 5)^4)^(1/3) == 625, x, Reals]
Solve[((6 x - 5)^(1/3))^4 == 625, x, Reals]
5

the standard definition of fractional roots is to use the denominator first. So (-20)^(4/3) is the same as ((-20)^(1/3))^4. Below I show the outcomes of the two possible groupings and one can see which agrees with the ungrouped form.

In[571]:= (-20)^(4/3)

Out[571]= -20 (-5)^(1/3) 2^(2/3)

In[572]:= ((-20)^4)^(1/3)

Out[572]= 20 2^(2/3) 5^(1/3)

In[573]:= ((-20)^(1/3))^4

Out[573]= -20 (-5)^(1/3) 2^(2/3)
6

Thank you for the replies, and forgive me, but I have no idea how to use Mathematica in its full programmable sense. I am actually more of a paper and pencil type for math anyhow (oh how I loved those 20 page proofs in school, handwritten all neat and pretty!)

Gianluca:
You said “I always warn my students that the familiar algebraic rules break down when dealing with powers with a negative base and noninteger exponent.”
Can you tell me what you teach after you warn them?

Also, I do understand how you can get imaginary numbers, but I am concerned with the other real answer of -20.
Thank you!

7

The rules we use to manipulate powers are true in two situations: any base but integer exponent, or positive base with any exponent. When you have a negative (or complex) base and noninteger exponent you cannot trust the rules. One simple example is with the cubic root of a negative number. We expect (-8)^(1/3) to be -2. But then we cannot replace 1/3 with 2/6, because (-8)^(2/6)=+2. I know of a “proof” that 0=1, where the error is hard to single out, unless we are aware of the problems with powers.