In all equations, do NOT set it equal to $x=$ or $y=$. If your answer is $x=5$, simply write in 5 as your answer.
$(2x)^{2}\cdot2x^{4}=$
${4x}^{6}$
${4x}^{8}$
${64x}^{6}$
${8x}^{6}$
$\frac{x^6}{x^{12}}=$
$\frac{1}{x^2}$
${x}^{18}$
$\frac{2}{x^2}$
$\frac{1}{x^6}$
$(\frac{1}{2})^{-3}=$
$-\frac{3}{2}$
$-8$
$8$
$\frac{1}{8}$
${25x}^{2}=(5x)^{2}$
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In the term ${-12x}^{3}$, $-12$ is the base.
$\sqrt{12}\sqrt{12}=12$
$\frac{5}{\sqrt5}=\sqrt{5}$
Solve for $x$: $3^{x}=1$
Solve for $x$: $(\frac{1}{3})^{x}=3$
Solve for $x$: $x^{3}=-8$
If $\sqrt{3}x=\sqrt{27}$, what is the value of $\frac{9}{7}x$?
Simplify: $5\sqrt{3}-\sqrt{3}$
Simplify: $(6\sqrt{5})^{2}$
Simplify so there are no negative exponents:
$\frac{18x^{3}y^{3}}{12x^{6}y^{2}}$
If $\sqrt{2x+3}-7=2$, what is $\frac{1}{3}{x}$?
Rationalize and simplify:
$\frac{12}{\sqrt{6}}$
Solve for $x$:
$\sqrt[3]{x}-4=0$