# textbook isn't helping me with this problem can anybody help

• Dec 11th 2010, 06:04 PM
bhuston3
textbook isn't helping me with this problem can anybody help
^5 square root of 5x-5 = ^5 square root of 6x-7

the ^ 5 is the little 5 on top before the square root symbol. I'm having trouble finding out how to solve for x
• Dec 11th 2010, 06:07 PM
Prove It
$\displaystyle \displaystyle \sqrt[5]{5x-5} = \sqrt[5]{6x - 7}$

$\displaystyle \displaystyle (\sqrt[5]{5x-5})^5 = (\sqrt[5]{6x-7})^5$

$\displaystyle \displaystyle 5x-5=6x-7$.

Go from here.
• Dec 11th 2010, 06:08 PM
TheCoffeeMachine
If $\displaystyle x^n = y^n$, then $\displaystyle x = y$, of course.
• Dec 11th 2010, 06:37 PM
snowtea
x^n = y^n sometimes has more solutions than just x = y.
Simplest example: x^2 = y^2 could be x = -y

Also, when raising to a power, we may actually have fewer solutions:
Consider sqrt(x) = -sqrt(y), squaring both sides gives x = y, but
sqrt(x) = -sqrt(x) is only true when x=0.
It is always a good idea to plug solutions back into the original equation to check.

For this problem, all solutions check out. :)
• Dec 11th 2010, 06:42 PM
Prove It
Quote:

Originally Posted by snowtea
x^n = y^n sometimes has more solutions than just x = y.

Simplest example: x^2 = y^2 could be x = -y

If $\displaystyle \displaystyle n$ is odd, this is true, at least for real solutions.