When solving probelms involving roots and exponents the following rules must taken into account.
For integers \(a\) and \(b\):
$$x^n\cdot x^m = x^{n+m},$$
$$\frac{x^n}{x^m} = x^{n-m},$$
$$\left(x^n\right)^m = x^{nm},$$
$$\left(xy\right)^m = x^m\cdot y^m,$$
$$\left(\frac{x}{y}\right)^m = \frac{x^m}{y^m}$$
The fractional exponent \(p/q\) applied on the variable \(x\) can be wrriten as: $$ x^{\frac{p}{q}} = \left(x^{1/q}\right)^p = \sqrt[q]{x^p}$$ where the value of \(x^{1/q}\) is positive qth root of \(x\) if the \(x > 0 \) and the negative qth root of \(x\) if the \(x\) is negative and \(q\) is odd.
Examples - Exponents and Roots
- Example 1 - Solve \(2^3 \cdot 2^4 \)
Solution $$ 2^3 \cdot 2^4 = 2^{3+4} =2^7$$ - Example 2 - Solve \(2^6:2^4\)
Solution $$ 2^6:2^4 = 2^{6-4} = 2^2.$$ - Example 3 - Solve \(2^2 \cdot 3 ^2. \)
Solution $$2^2 \cdot 3^2 = \left(2\cdot 3\right)^2 = 6^2 = 36. $$ - Example 4 - Solve \(6^2 : 3^2.\)
Solution $$ 6^2 : 3^2 = \left(6:3\right)^2 = 2^2 = 4 $$ - Example 5 - Solve \(\left(2^3\right)^2.\)
Solution $$ \left(2^3\right)^2 = 2^{2\cdot 3} = 2^6 = 64.$$ - Example 6 - Calcualte \(\left(-\frac{x^2 y^{-3}}{x^{-1}}\right)^{-3}. \)
Solution $$ \left(-\frac{x^2 y^{-3}}{x^{-1}}\right)^{-3} = \left(-x^3 y^{-3}\right)^{-3} = -x^{3\cdot(-3)} y^{(-3)\cdot (-3)} = -x^{-9}y^9 = -(\frac{y}{x})^9$$ - Example 7 - Calculate \(\frac{a^{x+1}}{b^{x-2}}:\left(\frac{a^{2x-1}}{b^{x+2}} \cdot \frac{a^{3x}}{b^6}\right)\)
Solution $$ \frac{a^{x+1}}{b^{x-2}}:\left(\frac{a^{2x-1}}{b^{x+2}} \cdot \frac{a^{3x}}{b^6}\right) = \frac{a^{x+1}}{b^{x-2}} : \frac{a^{5x-1}}{b^{x+8}} = \frac{a^{x+1}}{b^{x-2}} \cdot \frac{b^{x+8}}{a^{5x-1}} =$$ $$ = a^{x+1-5x+1} \cdot b^{x+8-x+2} = a^{2-4x} \cdot b^{10} $$ - Example 8 - Calculate \(\left[6:\left(4\cdot 5^2 - 7\cdot 13 -2^3\right)\right]^2.\)
Solution $$ \left[6:\left(4\cdot 5^2 - 7\cdot 13 -2^3\right)\right]^2 = \left[6:\left(100 - 91 - 8\right)\right]^2 = \left[6:1\right]^2 = 6^2 = 36.$$ - Example 9 - Calculate \(\frac{4a^3}{ab^{-4}} : \frac{2a^{-2}}{b^{-3}}.\)
Solution $$\frac{4a^3}{ab^{-4}} : \frac{2a^{-2}}{b^{-3}} = \frac{4a^2}{b^{-4}} \cdot \frac{b^{-3}}{2a^{-2}} = 2\cdot a^2 \cdot b^4 \cdot b^{-3}\cdot a^{2} = 2 \cdot a^4 \cdot b$$ - Example 10 - Calculate \(\left[\left(\frac{1}{3}\right)^{-3} \cdot 6^{-3}+ \left(\frac{1}{5}\right)^{-2}\cdot 2.5^{-2}\right]\cdot\left(-\frac{1}{2^3}\right)^2\)
Solution $$\left[\left(\frac{1}{3}\right)^{-3} \cdot 6^{-3}+ \left(\frac{1}{5}\right)^{-2}\cdot 2.5^{-2}\right]\cdot\left(-\frac{1}{2^3}\right)^2 = \left[27\cdot \frac{1}{216}+ \left(\frac{1}{5}\cdot \frac{5}{2}\right)^{-2}\right]\cdot\left(-\frac{1}{2^6}\right) =$$ $$ = \left(\frac{1}{8}+\frac{1}{2^{-2}}\right)\cdot\left(-\frac{1}{2^{6}}\right) = 2^{-3-6} + \frac{1}{2^{-2+6}} = 2^{-9} + 2^{-4} = \frac{1}{512} + \frac{1}{16} = \frac{33}{512}$$ - Example 11 - Calculate \(\left(9x^2y3-12x^4y^4\right):3x^2y - (2+3x^2y)\cdot y^2\).
Solution $$ \left(9x^2y3-12x^4y^4\right):3x^2y - (2+3x^2y)\cdot y^2 = 3x^{2-2}y^{3-1} - 4x^{4-2}y^{4-1} - (2y^2 + 3x^2 y^3) = $$ $$3y^2 - 4x^2y^3 - 2y^2 -3x^2y^3 = y^2 - 7x^2 y^3 = y^2\cdot(1-7x^2y)$$ - Example 12 - Determine which one of the numbers is bigger: \(9^{60}\) or \(27^{40}\)?
Solution $$9^{60} = \left(3^2\right)^{60} = 3^{2\cdot 60} = 3^{120}, $$ $$ 27^{40}=\left(3^3\right)^{40} = 3^{3\cdot 40} = 3^{120}$$ Answer: The \(9^{60}\) and \(27^{40}\) are equal. - Example 13 - Calculate 20% of \(\left(\frac{1}{2} - \frac{2}{5}\right)\left(16 + \frac{36}{4}\right)^{\frac{1}{2}}\).
Solution Before calculating 20% of the expression the expression must be solved. \begin{eqnarray}\left(\frac{1}{2} - \frac{2}{5}\right)\left(16 + \frac{36}{4}\right)^{\frac{1}{2}} &=& \\ \frac{1}{10}\left(\frac{64+36}{4}\right)^{\frac{1}{2}} &=& \\ \frac{1}{10}\left(\frac{100}{4}\right)^{\frac{1}{2}} &=& \\ \frac{1}{10}\frac{10}{2} &=& \frac{1}{2}\end{eqnarray} Now that the value of the expression is obtained the 20% can easily be calculated. $$ \frac{2}{10}\frac{1}{2} = \frac{1}{10}$$ - Example 14 - Calculate: \(\sqrt{\sqrt[3]{8}}\sqrt[4]{\sqrt[3]{x^2}}\)
Solution $$ \sqrt{\sqrt[3]{8}}\sqrt[4]{\sqrt[3]{x^2}} = \sqrt[6]{x^8} \cdot \sqrt[6]{x} = \sqrt[6]{x^9} = \sqrt{x^3} = x\sqrt{x}.$$ - Example 15 - Calculate: \(3\sqrt[3]{a^2\cdot \sqrt[4]{a}} + \sqrt{a\cdot\sqrt[6]{a^3}} - 2\sqrt[6]{a^3 \cdot \sqrt{a^3}}\)
Solution $$ 3\sqrt[3]{a^2\cdot \sqrt[4]{a}} + \sqrt{a\cdot\sqrt[6]{a^3}} - 2\sqrt[6]{a^3 \cdot \sqrt{a^3}} = $$ $$3\sqrt[12]{a^9} +\sqrt[6]{a^6\cdot a^3} - 2 \sqrt[12]{a^6\cdot a^3} = $$ $$3\sqrt[12]{a^9} + \sqrt[12]{a^9} - 2 \sqrt[12]{a^9} = 2 \sqrt[4]{a^3}.$$ - Example 16 - Calculate: \(\left[\left(625^{0.75}\right)^{-\frac{2}{3}}\cdot \left(8^{\frac{1}{3}}\right)^{-2}\right]^{-\frac{1}{2}}\)
Solution $$ \left[\left(625^{0.75}\right)^{-\frac{2}{3}}\cdot \left(8^{\frac{1}{3}}\right)^{-2}\right]^{-\frac{1}{2}} = \left[625^{-\frac{1}{2}}\cdot 2^{-2}\right]^{-\frac{1}{2}} = \left[5^{-2}\cdot 2^{-2}\right]^{-\frac{1}{2}} = \left[10^{-2}\right]^{-\frac{1}{2}} = 10$$ - Example 17 Calculate \(\left(2\sqrt{2}-3\sqrt{3}+5\sqrt{27}\right)\sqrt{6}\)
Solution $$\left(2\sqrt{2}-3\sqrt{3}+5\sqrt{27}\right)\sqrt{6} =2 \cdot 2\sqrt{2} - 3\cdot 3 \sqrt{2} + 5 \cdot 3\cdot 3\sqrt{2} = 4\sqrt{3}- 9\sqrt{2} + 45 \sqrt{2} = 4\sqrt{3} + 36\sqrt{2}$$ - Example 18 - Calculate: \((a+1)\cdot \sqrt{\frac{1}{a}- \frac{1}{a^2 +a }}\)
Solution $$ (a+1)\cdot \sqrt{\frac{1}{a}- \frac{1}{a^2 +a }} = (a+1) \cdot \sqrt{\frac{a+1-1}{a(a+1)}} = \sqrt{(a+1)^2 \cdot \frac{a}{a(a+1)}} = \sqrt{a+1}$$ - Example 19 - Calculate \(10\cdot 10^{n+2}.\)
Solution $$ 10\cdot 10^{n+2} = 10^{n+3}$$ - Example 20 - Calculate \(\frac{0.0001^2}{100\cdot 0.01}.\)
Solution $$ \frac{\left(10^{-4}\right)^2}{10^2 \cdot 10^{-2}} = \frac{10^{-4\cdot 2}}{10^{2-2}} = 10^{-8} = \frac{1}{10^8}.$$ - Example 21 - Simplify the following expression \(\left(\frac{x^{-3}}{y}\right)^{-2}\cdot \frac{x^3}{y^2}\)
Solution $$ \left(\frac{x^{-3}}{y}\right)^{-2}\cdot \frac{x^3}{y^2} = \frac{x^{-3\cdot (-2)}}{y^{-3}}\cdot \frac{x^3}{y^2} = x^6\cdot y^3\cdot x^{3}\cdot y^{-2} = x^{6+3}\cdot y^{3-2} = x^9\cdot y$$ - Example 22 - If \(n=10\) calculate the expression \(10^{n-7}\cdot \left(10+\frac{100}{10}\right)^{n-8}\)
Solution $$ 10^{n-7}\cdot \left(10+\frac{100}{10}\right)^{n-8} = 10^{3}\cdot \left(10+10\right)^{2} = 1000\cdot 400 = 400000$$ - Example 23 - The mass of Jupiter is equal to \(2\cdot 10^{27}\) kg while the mass of the Earth is \(6\cdot 10^{24}\) kg. Calculate how much the mass of Jupiter is larger than the mass of the Earth.
Solution $$ \frac{2\cdot 10^{27}}{6\cdot 10^{24}} = \frac{1}{3} \cdot 10^{27-24} = \frac{1000}{3} = 333.333$$ - Example 24 - The number 0.000002342 is equal to ? $$ 0.2345 = 2.345\cdot 10^{-6}$$
- Example 25 - The 100 \(\left[\mathrm{m}^2\right]\) is equla to how much \(\mathrm{cm}^2\) ?
Solution - 1 [m] is equal to 100 [cm] so, the $$1 \left[\mathrm{m}^2\right] = 100 [\mathrm{cm}] \cdot 100 [\mathrm{cm}] = 10^4 \left[\mathrm{m}^2\right]$$
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