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Learning Outcomes
- Exponents and Exponent laws with whole-number exponents
- Includes variable bases
- 2^7=2\times2\times2\times2\times2\times2\times2=128;
n^4=n\times n\times n\times n - Exponent laws ex. 6^0=1; m^1=m; n^5\times n^3=n^8; \frac{y^7}{y^3}=y^4;
(5n)^3=5^3\times n^3; \left(\frac{m}{n}\right)^5=\frac{m^5}{n^5}; and \left(3^2\right)^4=3^8 - Limited to whole-number exponents and whole-number exponent outcomes when simplified
- (-3)^2 does not equal -3^2
Lesson Examples
- 2^6
Solution64 - 3^4
Solution81 - Simplify a\times a\times a\times a\times a
Solutiona^5 - (-3)^4
Solution81 - n\times n^7\times n^4
Solutionn^{12} - \left(x^5\right)(x)\cdot x^3
Solutionx^9 - \left(3x^3\right)^3
Solution27x^9 - \left(\frac{2}{-3}\right)^3
Solution-\frac{8}{27} - Simplify \frac{x^5}{x^3}
Solutionx^2 - x^a\cdot x^b
Solutionx^{a+b} - \left(\frac{2}{3}\right)^3
Solution\frac{8}{27} - \left(\frac{a}{b}\right)^c
Solution\frac{a^c}{b^c} - Evaluate -5^2
Solution-25 - Evaluate (-5)^2
Solution25 - \frac{x^5}{x^3}\div \frac{x}{x^2}
Solutionx^3 - \frac{5ab^3c}{15a^3b^5}
Solution\frac{c}{3a^2b^2} - \left(x^2\right)^3
Solutionx^6 - \left(a^b\right)^c
Solutiona^{bc} - \left( \frac{2x^2y^3}{z^5} \right) ^3
Solution\frac{8x^6y^9}{z^{15}} - \frac{(-3)^4}{(-3)^2}\times -3^2
Solution-81 - -2(-2)^2-(-3)^2
Solution-17 - \left(\left(2x^2\right)^3\right)^2
Solution64x^{12} - \left(2\left(3a^2\right)^3\right)^2
Solution2916a^{12} - (-1)^{2024}
Solution1 - \frac{(-2)^{100}}{-2^{98}}
Solution-4 - 1^0+0^1
Solution1 - 0^0
SolutionUndefined - \left(\frac{4p^4p^6q^4}{2p^8q^3}\right)^3
Solution8p^6q^3 - \left(\frac{(-3)^2}{(-2)^4}\right)^2
Solution\frac{81}{256} - -2\left(-\frac{3}{4}\right)^2+(-1)^3-\left(\frac{-2^4}{(-2)^3}\right)^2
Solution-\frac{49}{8} - \frac{2}{x}\left(\frac{3x}{2x^2}\right)^3\div \frac{1}{x^3}
Solution\frac{27}{4x} - Solve 27=9^{3x}
Solutionx=\frac{1}{2} - Solve \frac{3^{10}}{3^x}=3^{x+1}
Solutionx=\frac{9}{2} - Solve 8^{1-2x}=\frac{2^9}{2^x}
Solutionx=-\frac{6}{5} - Challenge:
- \left[\frac{(-3^3)}{3^2}\right]^2-\left[\frac{(-3)^4}{3^3}\right]^1+\left[\frac{(-3)^1}{3}\right]^0
Solution7 - \left(\frac{8^{1-2x}}{2^{x+3}}\right)^4=2
Solutionx=-\frac{1}{28} - Solve 32\times 2+2^{x+1}=3\times 2^x
Solutionx=6
- \left[\frac{(-3^3)}{3^2}\right]^2-\left[\frac{(-3)^4}{3^3}\right]^1+\left[\frac{(-3)^1}{3}\right]^0
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