1. square root question.

$\sqrt{9^{16x^2}}=\begin{cases}\text{a) }3^{4x} &\text{b) } 3^{16x^2}\\\text{c) }9^{4x^2}&\text{d) }9^{8x}\end{cases}$

2. Re: square root question.

Spoiler:
I say (b) ...

3. Re: square root question.

Originally Posted by skeeter
Spoiler:
I say (b) ...
i agree

4. Re: square root question.

$\sqrt{9^{16x^2}} \ = \ \sqrt{(3^2)^{16x^2}} \ = \ \sqrt{3^{[2(16x^2)]}} \ = \ \sqrt{(3^{16x^2})^2} \ = \ 3^{16x^2}$ **

** $\sqrt{A^2} \ = \ |A|,$ but the expression for A here is always non-negative, so the final expression in the above line follows.

5. Re: square root question.

agree with you

6. Re: square root question.

I almost was fooled by (d)! (b) $3^{16x^2}$ is equivalent to $9^{8x^2}$ but (d), of course, does not have x squared.

7. Re: square root question.

Next, we can do some superroot problems! Inverse tetration is a hoot!