Question

Hi would appreciate some assistance with this. I am supposed to simplify and find an extraneous solution (if there is one).

(7a^2 b^3)^1/2 (7a^2b^3)^-1/2

Answer 1

Irish, can you explain...

Answer 2

Is that multiplication between the two expressions?

@dawidskalkowski

Answer 3

I just did the math, and I also got the answer 1. I'm just double-chcking. And yes genius, it is.

Answer 4

\((7a^2 b^3)^\frac{1}{2} (7a^2b^3)^{\frac{-1}{2}} \)

Note : \(a^{-m} = \cfrac{1}{a^m} \)

So, \(\large { (7a^2 b^3 ) ^\frac{-1}{2} = \cfrac{1}{(7a^2 b^3 ) ^\frac{1}{2}} }\)

Now, we get :
\(\large {(7a^2 b^3)^\frac{1}{2} (7a^2b^3)^{\frac{-1}{2}} \implies (7a^2 b^3 ) ^\frac{1}{2} \times \cfrac{1}{(7a^2 b^3 ) ^\frac{1}{2}} } \)

\(\large{\cancel{(7a^2 b^3) ^\frac{1}{2}} \times \cfrac{1}{\cancel{(7a^2 b^3 ) ^\frac{1}{2}}} }\) = \(\color{blue}{\large{1}} \)

Answer 5

that looks good

basically X ^(1/2) * X ^ (-1/2) = X ^ 0 = 1

Answer 6

Right, it can also be explained further in the form of identity :

\(\large a^m \times a^n = a^{m+n}\)

If m = -n

It becomes :

\(\large a^{\color{blue}{-n}} \times a^{n} = a^{\color{blue}{-n } + n }= a^0 = 1\)