# Math Help - Need help figuring out index law question .. stuck .. please help

1. ## Need help figuring out index law question .. stuck .. please help

hey guys theres this one question ive been working on forever ..

its an index law question

it asks to simplify it

i understand anything rooted is to the power of a half and i am absolutely clueless as to how to advance in this question .. everything else i can do fin except this one question

thanks guys ..

BTW the back of book answer is : x(x-1)^-0.5

2. Originally Posted by jimzer
hey guys

ive been working on this question for more than a day and i can do all others except this single one ..

simplify : <refer to attachment>

i understand anything rooted is to the power of negative 1/2 but i seriously cannot figure this question out ..

the back of book answers is : x(x-1)^0.5

thanks a lot guys
Get a common denominator:

$\frac{1}{\sqrt{x-1}} + \sqrt{x-1} = \frac{1}{\sqrt{x-1}} + \frac{\sqrt{x-1} \cdot \sqrt{x-1}}{\sqrt{x-1}}$

$\frac{1}{\sqrt{x-1}} + \frac{x-1}{\sqrt{x-1}} = \frac{1 + (x-1)}{\sqrt{x-1}}$

$= \frac{x}{\sqrt{x-1}} = \frac{x}{(x-1)^{1/2}} = x (x-1)^{-1/2}$.

The book's answer is wrong.

3. Originally Posted by mr fantastic
Get a common denominator:

$\frac{1}{\sqrt{x-1}} + \sqrt{x-1} = \frac{1}{\sqrt{x-1}} + \frac{\sqrt{x-1} \cdot \sqrt{x-1}}{\sqrt{x-1}}$

$\frac{1}{\sqrt{x-1}} + \frac{x-1}{\sqrt{x-1}} = \frac{1 + (x-1)}{\sqrt{x-1}}$

$= \frac{x}{\sqrt{x-1}} = \frac{x}{(x-1)^{1/2}} = x (x-1)^{-1/2}$.

The book's answer is wrong.
thank you so much man, just gotta figure it out now!!

and i wrote the wrong answer down its actually ^-0.5

thanks so much again, and sry for double posting