Solve the following equation. \( \log _{2} 8=x \) Rewrite the equation in exponential form. \( 8=2^{x} \) (Do not simplify. Type an equation. Type an exact answer, using The solution set is \( \mathbb{D}_{\text {I }} \). (Simplify your answer. Type an exact answer, using radicals as

1) \( \left(4 x^{2}+5\right)\left(x^{2}-2\right)\left(x^{2}+2\right)=0 \) A) \( \left\{\frac{i \sqrt{5}}{2},-\frac{i \sqrt{5}}{2}, \sqrt{2},-\sqrt{2}, i \sqrt{2},-i \sqrt{2}\right\} \) B) \( \left\{\frac{i \sqrt{10}}{2},-\frac{i \sqrt{10}}{2}, \sqrt{2},-\sqrt{2}, i \sqrt{2},-i \sqrt{2}\right\} \) C) \( \left\{\frac{i \sqrt{15}}{3},-\frac{i \sqrt{15}}{3}, i \sqrt{3},-i \sqrt{3}, i \sqrt{2},-i \sqrt{2}\right\} \) D) \( \left\{\right. \) mult. \( \left.2, \frac{i \sqrt{5}}{2},-\frac{i \sqrt{5}}{2}, \sqrt{2},-\sqrt{2}\right\} \)

Find the zeros for the polynomial function and give the multiplicity for each zero. State whether the graph crosses the \( x \)-axis or touches the \( x \)-axis and turns around at each zero. \( f(x)=x^{3}+3 x^{2}-16 x-48 \) Determine the behavior of the function at each zero. Select the correct choice below and, if necessary, fill in the answer boxes within your choice. A. The graph crosses the \( x \)-axis at \( x=\square \). The graph touches the \( x \)-axis and turns around at \( x=\square \). (Type integers or decimals. Simplify your answers. Use a comma to separate answers as needed.) B. The graph crosses the \( x \)-axis at all zeros. C. The graph touches the \( x \)-axis and turns around at all zeros.

dent Name: Which graph shows the solution of \( -6(n-1)<3 \) and \( 2(n+1)>0 \) ?

dation \( \log _{3}(2 x-1)=3 \)

Watch the video that describes performing vector operations. Click here to watch the video. Given vectors \( \mathbf{u} \) and \( \mathbf{v} \), find \( \mathbf{( a )} 9 \mathbf{u}=\langle 9,9\rangle, \mathbf{v}=\langle 6,0\rangle \) (b) \( 9 \mathbf{u}+6 \mathbf{v} \) (c) \( \mathbf{v}-6 \mathbf{u} \). \( \begin{array}{ll}\text { (a) } 9 \mathbf{u}=\langle 81,81\rangle \\ \text { (b) } 9 \mathbf{u}+6 \mathbf{v}=\langle\square, \square\rangle\end{array} \)