In this exercise you will practice with the exponent laws, all mixed-up.
These problems require the application of more than one exponent law.
For simpler problems, see One-Step Exponent Law Practice.
[beautiful math coming... please be patient] $\displaystyle\left(\frac{1}{x^2}\right)^3 = (x^{-2})^3 = x^{-2\,\cdot\, 3} = x^{-6} = x^p$ | where [beautiful math coming... please be patient] $\,p = -6$ |
[beautiful math coming... please be patient] $\displaystyle \left(\frac{x^2}{x^3}\right)^5 = (x^{2-3})^5 = (x^{-1})^5 = x^{-1\,\cdot\, 5} = x^{-5} = x^p $ | where [beautiful math coming... please be patient] $p = -5$ |
[beautiful math coming... please be patient] $(x^2x^4)^{-1} = (x^{2+4})^{-1} = (x^6)^{-1} = x^{6\,\cdot\, -1} = x^{-6} = x^p$ | where [beautiful math coming... please be patient] $\,p = -6$ |
[beautiful math coming... please be patient] $\displaystyle \frac{x^2x^{-3}}{x^5} = \frac{x^{2 + (-3)}}{x^5} = \frac{x^{-1}}{x^5} = x^{-1-5} = x^{-6} = x^p$ | where [beautiful math coming... please be patient] $\,p = -6$ |
[beautiful math coming... please be patient] $\displaystyle \frac{x^2}{x^3x^4} = \frac{x^2}{x^{3+4}} = \frac{x^2}{x^7} = x^{2-7} = x^{-5} = x^p$ | where [beautiful math coming... please be patient] $\,p = -5$ |
[beautiful math coming... please be patient] $\displaystyle \frac{(x^2)^3}{(x^{-1})^4} = \frac{x^{2\,\cdot\,3}}{x^{-1\,\cdot\,4}} = \frac{x^6}{x^{-4}} = x^{6-(-4)} = x^{10} = x^p$ | where [beautiful math coming... please be patient] $\,p = 10$ |