Solve for X (Expressions with Roots)

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mj.silverfly
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Joined: Fri May 27, 2016 9:52 pm
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Solve for X (Expressions with Roots)

I am working on a question and I have the solution to it. I just can't seem to grasp how the last step happens. Can someone please help explain it to me?

Solve for x:

ux^a = x

The answer they give is:

Step 1: ux^a = x

Step 2: x^(a-1) = (1/u)

Step 3: x = (1/u)^((1/(a-1))

I understand until that part. The next step, I don't understand.

Step 4: x = u^(1/(1-a))

How do you get from Step 3 to Step 4?

Thank you!

theshadow
Posts: 136
Joined: Sun Feb 22, 2009 11:12 pm

Re: Solve for X (Expressions with Roots)

Solve for x: ux^a = x

The answer they give is:

Step 1: ux^a = x

Step 2: x^(a-1) = (1/u)

Step 3: x = (1/u)^((1/(a-1))

I understand until that part. The next step, I don't understand.

Step 4: x = u^(1/(1-a))

How do you get from Step 3 to Step 4?
They used exponent rules and flipping a subtraction.

. . .$\dfrac{1}{u}\, =\, u^{-1}$

. . .$a\, -\, 1\, =\, -1\, +\, a\, =\, -(1\, -\, a)\, =\, -1(1\, -\, a)$

. . .\begin{align}\left(\dfrac{1}{u}\right)^{\left(\dfrac{1}{a\, -\, 1}\right)}\, &=\, \left(u^{-1}\right)^{\left(\dfrac{1}{a\, -\, 1}\right)}\, \\ \\ &=\, \left(u^{-1}\right)^{\left(\dfrac{1}{-1(1\, -\, a)}\right)}\, \\ \\ &=\, \left(u^{-1}\right)^{\left(\dfrac{-1}{+1(1\, -\, a)}\right)} \\ \\ &=\, u^{(-1)\, \left(\dfrac{-1}{(1\, -\, a)}\right)} \\ \\ &=\, u^{\left(\dfrac{(-1)\, (-1)}{1\, -\, a}\right)} \\ \\ &=\, u^{\left(\dfrac{+1}{1\, -\, a}\right)} \\ \\ &=\, u^{\left(\dfrac{1}{1\, -\, a}\right)} \end{align}

Hope that helps!

mj.silverfly
Posts: 3
Joined: Fri May 27, 2016 9:52 pm
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Re: Solve for X (Expressions with Roots)

Oh, wow. I get it now. I wouldn't have thought to go through all of that, though! I probably would have left it at Step 3. I guess it's more simplified after Step 4?

Thank you so much for your quick reply!

mj.silverfly
Posts: 3
Joined: Fri May 27, 2016 9:52 pm
Contact:

Re: Solve for X (Expressions with Roots)

Thank you, thank you!

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