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fractional exponents and find extraneous solutions in W|A?

M P
Posted 9 years ago

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!
POSTED BY: M P
6 Replies
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POSTED BY: Gianluca Gorni
M P
M P
Posted 9 years ago

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!
POSTED BY: M P

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)
POSTED BY: Daniel Lichtblau

Also, compare

Solve[((6 x - 5)^4)^(1/3) == 625, x, Reals]
Solve[((6 x - 5)^(1/3))^4 == 625, x, Reals]
POSTED BY: Gianluca Gorni
POSTED BY: Gianluca Gorni

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
POSTED BY: S M Blinder
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