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# End Of Chapter Exercises

Exercise 2.4

Simplify:

$$(\text{8}x)^{\text{3}}$$

\begin{align*} (\text{8} x) ^{\text{3}} &= \text{8} ^{\text{3}} x^{\text{3}}\\ &= \text{512} x ^{\text{3}} \end{align*}

$$t^{3} \times 2t^{0}$$

\begin{align*} t^{3} \times 2t^{0} & = t^{3} \times 2(1) \\ & = 2t^{3} \end{align*}

$$5^{2x + y} \times 5^{3(x + z)}$$

\begin{align*} 5^{2x + y} \times 5^{3(x + z)} & = 5^{2x + y + 3x + 3z} \\ & = 5^{5x + y + 3z} \end{align*}

$$\text{15}^{3x} \times \text{15}^{12x}$$

\begin{align*} \text{15}^{3x} \times \text{15}^{12x} &= \text{15} ^{3x + 12x }\\ &= \text{15} ^{15x} \end{align*}

$$\dfrac{\text{7}^{y+7}}{\text{7}^{y+6}}$$

$\frac{\text{7}^{ y +7}} {\text{7}^{ y +6}} = \text{7}^{(y +7) - (y +6)}$

\begin{align*} \text{7}^{(y+7) - (y+6)} &= \text{7}^{y+7 - y-6} \\ &= \text{7}^{1} \\ &= \text{7} \end{align*}

$$3(d^4)(7d^3)$$
$3(d^4)(7d^3) = 21d^7$
$$(\frac{1}{7}a^2b^9)(6a^6b^2)(-3a^7b)$$
$\left(\frac{1}{7}a^2b^9\right)(6a^6b^2)(-3a^7b)= -\frac{18}{7}a^{15}b^{12}$

$$\left(b^{k + 1}\right)^{k}$$

$\left(b^{k + 1}\right)^{k} = b^{k^{2} + k}$

$$\dfrac{\text{24} c^{\text{8}} m^{\text{7}}}{\text{6} c^{\text{2}} m^{\text{5}}}$$

\begin{align*} &\frac{\text{24}c^{\text{8}}m^{\text{7}}}{\text{6}c^{\text{2}}m^{\text{5}}}\\ &= \text{4}c^{(\text{8} - \text{2})}m^{(\text{7} - \text{5})} \\ &= \text{4}c^{\text{6}}m^{\text{2}} \end{align*}

$$\dfrac{2(x^4)^{3}}{x^{12}}$$
\begin{align*} \frac{2(x^4)^{3}}{x^{12}} & = \frac{2x^{12}}{x^{12}} \\ & = 2 \end{align*}
$$\dfrac{a^6b^5}{7(a^8b^3)^2}$$
\begin{align*} \frac{a^6b^5}{7(a^8b^3)^2} & = \frac{a^6b^5}{7a^{16}b^6} \\ & = \frac{1}{7a^{10}b} \end{align*}
$$\left(\frac{a^7}{b^4}\right)^2$$
$\left(\frac{a^7}{b^4}\right)^2 = \frac{a^{14}}{b^8}$

$$\dfrac{6^{5p}}{9^{p}}$$

\begin{align*} \frac{6^{5p}}{9^{p}} & = \frac{2^{5p}.3^{5p}}{3^{2p}}\\ & = 2^{5p}.3^{5p - 2p} \\ & = 2^{5p}.3^{3p} \end{align*}

$$m^{-2t} \times \left(3m^{t}\right)^{3}$$

\begin{align*} m^{-2t} \times \left(3m^{t}\right)^{3} & = m^{-2t} \times 3^{3}m^{3t}\\ & = m^{-2t + 3t}.27 \\ & = 27m^{t} \end{align*}

$$\dfrac{3x^{-3}}{\left(3x\right)^{2}}$$

\begin{align*} \frac{3x^{-3}}{\left(3x\right)^{2}} & = 3^{1 - 2}.x^{-3 - 2}\\ & = 3^{-1}.x^{-5} \\ & = \frac{1}{3x^{5}} \end{align*}

$$\dfrac{5^{b - 3}}{5^{b + 1}}$$

\begin{align*} \frac{5^{b - 3}}{5^{b + 1}} & = 5^{b - 3 - b - 1} \\ & = 5^{-4} \\ & = \frac{1}{625} \end{align*}

$$\dfrac{2^{a - 2}3^{a + 3}}{6^{a}}$$

\begin{align*} \frac{2^{a - 2}3^{a + 3}}{6^{a}} & = \frac{2^{a - 2}3^{a + 3}}{\left(2.3\right)^{a}} \\ & = \frac{2^{a - 2}3^{a + 3}}{2^{a}.3^{a}}\\ & = 2^{a - 2 - a}.3^{a + 3 - a} \\ & = 2^{-2}.3^{3} \\ & = \frac{27}{4} \end{align*}

$$\dfrac{3^{n}9^{n - 3}}{27^{n - 1}}$$

\begin{align*} \frac{3^{n}9^{n - 3}}{27^{n - 1}} & = \dfrac{3^{n}.\left(3^{2}\right)^{n - 3}}{\left(3^{3}\right)^{n - 1}} \\ & = \dfrac{3^{n}.3^{2n - 6}}{3^{3n - 3}}\\ & = 3^{n + 2n - 6 - 3n + 3} \\ & = 3^{-3} \\ & = \frac{1}{27} \end{align*}
$$\dfrac{3^3}{9^3}$$
\begin{align*} \frac{3^3}{9^3} & = \left(\frac{3}{9}\right)^3 \\ & = \frac{1}{3^3} \\ & = \frac{1}{27} \end{align*}
$$\dfrac{x^{-1}}{x^4y^{-2}}$$
$\frac{x^{-1}}{x^4 y^{-2}} = \frac{y^{2}}{x^{5}}$
$$\dfrac{(-1)^4}{(-2)^{-3}}$$
\begin{align*} \frac{(-1)^4}{(-2)^{-3}} & = \frac{1}{(-2)^{-3}} \\ & = (-2)^3 \\ & = -8 \end{align*}

$$\left(\dfrac{2x^{2a}}{y^{-b}}\right)^{3}$$

\begin{align*} \left(\dfrac{2x^{2a}}{y^{-b}}\right)^{3} & = \dfrac{2^{3}\left(x^{2a}\right)^{3}}{\left(y^{-b}\right)^{3}} \\ & = \dfrac{2^{3}x^{6a}}{y^{-3b}}\\ & = 2^{3}x^{6a}y^{3b} \\ & = 8x^{6a}y^{3b} \end{align*}

$$\dfrac{2^{3x - 1}8^{x + 1}}{4^{2x - 2}}$$

\begin{align*} \dfrac{2^{3x - 1}8^{x + 1}}{4^{2x - 2}} & = \dfrac{2^{3x - 1}.2^{3(x + 1)}}{2^{2(2x - 2)}} \\ & = 2^{3x - 1 + 3x + 3 - 4x + 4}\\ & = 2^{2x + 6}\\ & = 4^{x + 3} \end{align*}

$$\dfrac{6^{2x}11^{2x}}{22^{2x - 1}3^{2x}}$$

\begin{align*} \frac{6^{2x}11^{2x}}{22^{2x - 1}3^{2x}} & = \frac{\left(3 \cdot 2\right)^{2x} \cdot 11^{2x}}{\left(2 \cdot 11\right)^{2x - 1} \cdot 3^{2x}} \\ & = \frac{3^{2x} \cdot 2^{2x} \cdot 11^{2x}}{2^{2x - 1} \cdot 11^{2x - 1} \cdot 3^{2x}} \\ & = 3^{2x - 2x} \cdot 2^{2x - 2x + 1} \cdot 11^{2x - 2x + 1} \\ & = 3^{0} \cdot 2^{1} \cdot 11^{1} \\ & = 22 \end{align*}

$$\dfrac{\left(-3\right)^{-3}\left(-3\right)^{2}}{\left(-3\right)^{-4}}$$

\begin{align*} \dfrac{\left(-3\right)^{-3}\left(-3\right)^{2}}{\left(-3\right)^{-4}} & = \left(-3\right)^{-3 + 2 + 4} \\ & = \left(-3\right)^{3} \\ & = -27 \end{align*}

$$\left(3^{-1} + 2^{-1}\right)^{-1}$$

\begin{align*} \left(3^{-1} + 2^{-1}\right)^{-1} & = \left(\dfrac{1}{3} + \dfrac{1}{2}\right)^{-1} \\ & = \left(\dfrac{2}{6} + \dfrac{3}{6}\right)^{-1} \\ & = \left(\dfrac{5}{6}\right)^{-1} \\ & = \dfrac{5^{-1}}{6^{-1}} \\ & = \dfrac{6}{5} \end{align*}

Simplify:

$$\dfrac{9^{n - 1}.27^{3 - 2n}}{81^{2 - n}}$$

\begin{align*} \dfrac{9^{n - 1}.27^{3 - 2n}}{81^{2 - n}} & = \dfrac{3^{2(n - 1)}.3^{3(3 - 2n)}}{3^{4(2 - n)}} \\ & = 3^{2(n - 1) + 3(3 - 2n) - 4(2 - n)} \\ & = 3^{2n - 2 + 9 - 6n - 8 + 4n} \\ & = \dfrac{1}{3} \end{align*}

$$\dfrac{2^{3n + 2} \cdot 8^{n - 3}}{4^{3n - 2}}$$

\begin{align*} \frac{2^{3n + 2} \cdot 8^{n - 3}}{4^{3n - 2}} & = \frac{2^{3n + 2} \cdot 2^{3(n - 3)}}{2^{2(3n - 2)}} \\ & = 2^{3n + 2 + 3(n - 3) - 2(3n - 2)} \\ & = 2^{3n + 2 + 3n - 9 - 6n + 4} \\ & = \dfrac{1}{8} \end{align*}

$$\dfrac{3^{t + 3} + 3^{t}}{2 \times 3^{t}}$$

\begin{align*} \dfrac{3^{t + 3} + 3^{t}}{2 \times 3^{t}} & = \dfrac{3^{t}.3^{3} + 3^{t}}{2.3^{t}} \\ & = \dfrac{3^{t}(3^{3} + 1)}{2.3^{t}} \\ & = \dfrac{3^{3} + 1}{2} \\ & = \dfrac{28}{2} \\ & = 14 \end{align*}

$$\dfrac{2^{3p} + 1}{2^{p} + 1}$$

\begin{align*} \dfrac{2^{3p} + 1}{2^{p} + 1} & = \dfrac{\left(2^{p} + 1\right)\left(2^{2p} - 2^{p} + 1\right)}{\left(2^{p} + 1\right)} \\ & = 2^{2p} - 2^{p} + 1 \end{align*}
$$\left(a^{10}b^{6}\right)^{\frac{1}{2}}$$
$\left(a^{10}b^{6}\right)^{\frac{1}{2}} = a^5b^3$
$$\left(9x^8y^4\right)^{\frac{1}{2}}$$
$\left(9x^8y^4\right)^{\frac{1}{2}} = 3x^4y^2$

$$\dfrac{\text{13}^{a}+ \text{13}^{a+2}}{\text{6} \times \text{13}^{a}- \text{13}^{a}}$$

\begin{align*} \frac{\text{13}^{a}+ \text{13}^{a+2}}{\text{6} \times \text{13}^{a}- \text{13}^{a}} &= \frac{\text{13}^{a}(\text{1} +\text{13}^{2})}{\text{13}^{a}(\text{6} -\text{1})} \\ &= \frac{(\text{1} +\text{13}^{2})}{(\text{6} -\text{1})} \\ &= \frac{\text{1} +\text{169}}{\text{6} -\text{1}} \\ &= \frac{\text{170}}{\text{5}} \\ &= \frac{\text{34}}{\text{1}} \\ & = \text{34} \end{align*}

$$\dfrac{\text{3}^{\text{8}z} \times \text{27}^{\text{8}z} \times \text{3}^\text{2}}{\text{9}^{\text{6}z}}$$

\begin{align*} \frac{\text{3}^{\text{8} z} \times \text{27} ^{\text{8} z} \times \text{3} ^\text{2}}{\text{9} ^{\text{6} z}} & = \frac{\text{3} ^{\text{8} z} \times {\left(\text{3} ^\text{3} \right) }^{\text{8} z} \times \text{3} ^\text{2}}{{\left( \text{3} ^\text{2} \right)} ^{\text{6} z}} \\ & = \frac{ \text{3} ^{\text{8} z} \times \text{3} ^{\text{24} z} \times \text{3}^\text{2}}{\text{3}^{\text{12}z}}\\ &= \frac{\text{3}^{\text{8}z + \text{24}z + \text{2}}}{\text{3}^{\text{12}z}}\\ &= \frac{\text{3}^{\text{32}z + \text{2}}}{\text{3}^{\text{12}z}}\\ &= \text{3}^{\text{32}z + \text{2} - \text{12}z }\\ &= \text{3}^{\text{20}z + \text{2}} \end{align*}

$$\dfrac{\text{121}^{b}- \text{16}^{p}}{\text{11}^{b}+ \text{4}^{p}}$$

\begin{align*} \frac{ \text{121} ^{b} - \text{16} ^{p}}{\text{11} ^{b}+ \text{4}^{p}} & = \frac{{\left(\text{11}^\text{2}\right)}^b- {\left(\text{4}^\text{2}\right)}^p}{\text{11}^{b}+ \text{4}^{p}} \\ &= \frac{{\left(\text{11}^b\right)}^{\text{2}}- {\left(\text{4}^p\right)}^{\text{2}}}{\text{11}^{b}+ \text{4}^{p}} \\ &= \frac{\left(\text{11}^b- \text{4}^p\right)\left(\text{11}^b+ \text{4}^p\right)}{\text{11}^{b}+ \text{4}^{p}} \\ &= \frac{\left(\text{11}^b- \text{4}^p\right)\left(\text{11}^b+ \text{4}^p\right)}{\text{11}^{b}+ \text{4}^{p}} \\ &= \text{11}^{b}- \text{4}^{p} \end{align*}

$$\dfrac{\text{11}^{-\text{4}c-4} \text{4}^{\text{4}c-3}}{\text{22}^{-\text{6}c-2}}$$

\begin{align*} \frac{\text{11} ^{-4c -4} \text{4} ^{4c -3}}{\text{22} ^{-6c -2}} & = \frac{\text{11}^{-4c-4} {\left( \text{2} ^\text{2} \right)} ^{4 c -3}} {{ \left( \text{11} \times \text{2} \right)} ^{-6 c-2}} \\ &= \frac{\text{11}^{-4c-4} \text{2}^{8c-6}} {\text{11}^{-6c-2}\text{2}^{-6c-2}}\\ &= \text{11}^{(-4c-4) - (-6c-2)} \times \text{2}^{(8c-6) - (-6c-2)}\\ &= \text{11}^{2c -\text{2}} \times \text{2}^{14c-\text{4}} \end{align*}

$$\dfrac{12^4\times 2^4}{16^6 \times 10}$$
\begin{align*} \frac{12^4\times 2^4}{16^6 \times 10} & = \frac{(3 \times 2^{2})^{4} \times 2^{4}}{(2^{4})^{6} \times (2 \times 5)}\\ & \frac{3^{4} \times 2^{8} \times 2^{4}}{2^{24} \times 2 \times 5} \\ & = \frac{3^{4}}{2^{13} \times 5} \end{align*}
$$\dfrac{5^6\times 3^{16} \times 2^7}{10^8 \times 9^6}$$
\begin{align*} \frac{5^6\times 3^{16} \times 2^7}{10^8 \times 9^6} & = \frac{5^6 3^{16} 2^7}{2^8 5^8 3^{12}} \\ & = \frac{3^4}{2 \times 5^2} \\ & = \frac{81}{50} \end{align*}

$$(\text{0,81})^ {\frac{\text{1}}{\text{2}}}$$

\begin{align*} (\text{0,81})^ {\frac{\text{1}}{\text{2}}} &= \left(\frac{\text{81}}{\text{100}}\right)^{\frac{1}{2}} \\ &= \left(\frac{\text{9}^\text{2}}{\text{10}^\text{2}}\right)^{\frac{1}{2}} \\ &= \left[\left( \frac{\text{9}}{\text{10}} \right)^\text{2} \right]^{\frac{1}{2}} \\ &= \frac{\text{9}}{\text{10}} \end{align*}

$$\text{12} {\left( a^\text{10}b^\text{20} \right)}^ {\frac{\text{1}}{\text{5}}} \times {\left( \text{729}a^\text{12}b^\text{15} \right)}^ {\frac{\text{1}}{\text{3}}}$$

\begin{align*} \text{12} {\left( a^\text{10}b^\text{20} \right)}^ {\frac{\text{1}}{\text{5}}} \times {\left( \text{729}a^\text{12}b^\text{15} \right)}^ {\frac{\text{1}}{\text{3}}} &= \text{12} a^{\frac{\text{10}}{\text{5}}}b^ {\frac{\text{20}}{\text{5}}} \times (\text{729})^{\frac{\text{1}}{\text{3}}}a^{\frac{\text{12}}{\text{3}}}b^{\frac{\text{15}}{\text{3}}} \\ &= \text{12} a^{\text{2}}b^{\text{4}} \times \left( \text{9}^{\text{3}} \right) ^{\frac{\text{1}}{\text{3}}}a^{\text{4}}b^{\text{5}} \\ &= \text{12} a^{\text{2}}b^{\text{4}} \times \text{9}a^{\text{4}}b^{\text{5}} \\ &= \text{108} a^{\text{6}} b^{\text{9}} \end{align*}

$$\text{2} {\left( p^\text{30}q^\text{20} \right)}^ {\frac{\text{1}}{\text{5}}} \times {\left( \text{1 331}p^\text{12}q^\text{6} \right)}^ {\frac{\text{1}}{\text{3}}}$$

\begin{align*} \text{2} {\left( p^\text{30}q^\text{20} \right)}^ {\frac{\text{1}}{\text{5}}} \times {\left( \text{1 331}p^\text{12}q^\text{6} \right)}^ {\frac{\text{1}}{\text{3}}} &= \text{2} p^{\frac{\text{30}}{\text{5}}}q^ {\frac{\text{20}}{\text{5}}} \times (\text{1 331})^{\frac{\text{1}}{\text{3}}}p^{\frac{\text{12}}{\text{3}}}q^{\frac{\text{6}}{\text{3}}} \\ &= \text{2} p^{\text{6}}q^{\text{4}} \times \left( \text{11}^{\text{3}} \right) ^{\frac{\text{1}}{\text{3}}}p^{\text{4}}q^{\text{2}} \\ &= \text{2} p^{\text{6}}q^{\text{4}} \times \text{11}p^{\text{4}}q^{\text{2}} \\ &= \text{22} p^{\text{10}} q^{\text{6}} \end{align*}

$$\dfrac{a^{-1}-b^{-1}}{a-b}$$
\begin{align*} \frac{a^{-1}-b^{-1}}{a-b} &= \frac{\frac{1}{a}-\frac{1}{b}}{a-b} \\ &= \frac{\frac{b-a}{ab}}{a-b} \\ &= \frac{-(a-b)}{ab(a-b)} \\ &= -\frac{1}{ab} \\ &= -(ab)^{-1} \end{align*}
$$\left(\left(x^{36}\right)^{\frac{1}{2}}\right)^{\frac{1}{3}}$$
\begin{align*} \left(\left(x^{36}\right)^{\frac{1}{2}}\right)^{\frac{1}{3}} & = \left(x^{18}\right)^{\frac{1}{3}} \\ & = x^6 \end{align*}
$$\left(\dfrac{2}{3}\right)^{x+y} \cdot \left(\dfrac{3}{2}\right)^{x-y}$$
\begin{align*} \left(\frac{2}{3}\right)^{x+y} \cdot \left(\frac{3}{2}\right)^{x-y} &= \left(\frac{2}{3}\right)^{x+y}\cdot \left(\frac{2}{3}\right)^{-(x-y)} \\ &=\left(\frac{2}{3}\right)^{x+y - (x-y)} \\ &=\left(\frac{2}{3}\right)^{2y} \end{align*}
$$(a^{\frac{1}{2}} + a^{-\frac{1}{2}})^2-(a^{\frac{1}{2}} - a^{-\frac{1}{2}})^2$$
\begin{align*} (a^{\frac{1}{2}} + a^{-\frac{1}{2}})^2-(a^{\frac{1}{2}} - a^{-\frac{1}{2}})^2 &= (a^{\frac{1}{2}} + a^{-\frac{1}{2}} - (a^{\frac{1}{2}} - a^{-\frac{1}{2}})) (a^{\frac{1}{2}} + a^{-\frac{1}{2}} + (a^{\frac{1}{2}} - a^{-\frac{1}{2}})) \\ &= (2a^{-\frac{1}{2}})(2a^{\frac{1}{2}})\\ &= 4a^{\frac{1}{2}-\frac{1}{2}}\\ &= 4a^{0} \\ &= 4 \end{align*}

Solve:

$$3^{x} = \dfrac{1}{27}$$

\begin{align*} 3^{x} & = \frac{1}{27} \\ 3^{x} & = \frac{1}{3^{3}} \\ 3^{x} & = 3^{-3} \\ \therefore x & = -3 \end{align*}

$$\text{121} = \text{11}^ {m -1}$$

\begin{align*} \text{121} & = \text{11}^ {m -1} \\ \text{11}^\text{2} & = \text{11}^ {m -1} \\ \therefore \text{2} & = m -1 \\ \text{2} +1 & = m \\ \text{3} & = m \end{align*}

$$5^{t - 1} = 1$$

\begin{align*} 5^{t - 1} & = 1 \\ 5^{t - 1} & = 5^{0} \\ \therefore t - 1 & = 0 \\ t & = 1 \end{align*}

$$2 \times 7^{3x} = 98$$

\begin{align*} 2 \times 7^{3x} & = 98 \\ 7^{3x} & = 49 \\ 7^{3x} & = 7^{2} \\ \therefore 3x & = 2 \\ x & = \dfrac{2}{3} \end{align*}

$$- \frac{64}{3} = - \frac{4}{3} 2^{- \frac{c}{3} + 1}$$

\begin{align*} \left( - \frac{3}{4} \right) \left( - \frac{64}{3} \right) & = \left( - \frac{4}{3} \cdot 2^{- \frac{c}{3} + 1} \right) \left( - \frac{3}{4} \right) \\ 16 & = 2^{- \frac{c}{3} + 1} \\ \therefore 2^{4} & = 2^{- \frac{c}{3} + 1} \\ 4 & = - \frac{c}{3} + 1 \\ -9 & = c \end{align*}

$$- \frac{1}{2} 6^{- n - 3} = -18$$

\begin{align*} \left( -2 \right) \left( - \frac{1}{2} \cdot 6^{- n - 3} \right) & = \left( -18 \right) \left( -2 \right) \\ 6^{- n - 3} & = 36 \\ 6^{- n - 3} & = 6^{2} \\ \therefore - n - 3 & = 2 \\ n & = -5 \end{align*}

$$2^{m + 1} = \left(\text{0,5}\right)^{m - 2}$$

\begin{align*} 2^{m + 1} & = \left(\text{0,5}\right)^{m - 2} \\ 2^{m + 1} & = \left(\dfrac{1}{2}\right)^{m - 2} \\ 2^{m + 1} & = \left(2^{-1}\right)^{m - 2} \\ 2^{m + 1} & = 2^{2 - m} \\ \therefore m + 1 & = 2 - m \\ m & = \dfrac{1}{2} \end{align*}

$$3^{y + 1} = 5^{y + 1}$$

\begin{align*} 3^{y + 1} & = 5^{y + 1} \\ \therefore y + 1 & = 0 \\ y & = -1 \end{align*}

$$z^{\frac{3}{2}} = 64$$

\begin{align*} z^{\frac{3}{2}} & = 64 \\ z^{\frac{3}{2}} & = 4^{3} \\ \left(z^{\frac{3}{2}}\right)^{\frac{2}{3}} & = \left(4^{3}\right)^{\frac{2}{3}} \\ z & = 4^{2} \\ z & = 16 \end{align*}

$$16x^{\frac{1}{2}} - 4 = 0$$

\begin{align*} 16x^{\frac{1}{2}} - 4 & = 0 \\ 16x^{\frac{1}{2}} &= 4 \\ x^{\frac{1}{2}} & = \dfrac{4}{16} \\ x^{\frac{1}{2}} & = \dfrac{1}{4} \\ \left(x^{\frac{1}{2}}\right)^{2} & = \left(\dfrac{1}{4}\right)^{2} \\ x & = \dfrac{1}{16} \end{align*}

$$m^{0} + m^{-1} = 0$$

\begin{align*} m^{0} + m^{-1} & = 0 \\ 1 + m^{-1} &= 0 \\ m^{-1} & = -1 \\ \left(m^{-1}\right)^{-1} & = \left(-1\right)^{-1} \\ m & = -1 \end{align*}

$$t^{\frac{1}{2}} - 3t^{\frac{1}{4}} + 2 = 0$$

\begin{align*} t^{\frac{1}{2}} - 3t^{\frac{1}{4}} + 2 & = 0 \\ \left(t^{\frac{1}{4}} - 1\right)\left(t^{\frac{1}{4}} - 2\right) &= 0 \\ t^{\frac{1}{4}} - 1 = 0 & \text{ or } t^{\frac{1}{4}} - 2 = 0 \\ t^{\frac{1}{4}} = 1 & \text{ or } t^{\frac{1}{4}} = 2 \\ \left(t^{\frac{1}{4}}\right)^{4} = (1)^{4} & \text{ or } \left(t^{\frac{1}{4}}\right)^{4} = (2)^{4} \\ t & = 1 \text{ or } 16 \end{align*}

$$3^{p} + 3^{p} + 3^{p} = 27$$

\begin{align*} 3^{p} + 3^{p} + 3^{p} & = 27 \\ 3.3^{p} & = 27 \\ 3^{p + 1} & = 3^{3} \\ \therefore p + 1 & = 3 \\ p & = 2 \end{align*}

$$k^{-1} - 7k^{-\frac{1}{2}} - 18 = 0$$

\begin{align*} k^{-1} - 7k^{-\frac{1}{2}} - 18 & = 0 \\ \left(k^{-\frac{1}{2}} - 9\right)\left(k^{-\frac{1}{2}} + 2\right) &= 0 \\ k^{-\frac{1}{2}} - 9 = 0 & \text{ or } k^{-\frac{1}{2}} + 2 = 0 \\ k^{-\frac{1}{2}} = 9 & \text{ or } k^{-\frac{1}{2}} = -2 \\ \left(k^{-\frac{1}{2}}\right)^{-2} = (9)^{-2} & \text{ or } \left(k^{-\frac{1}{2}}\right)^{-2} = (-2)^{-2} \\ k & = \frac{1}{81} \text{ or } \frac{1}{4} \end{align*}

We check both answers and find that $$k = \frac{1}{81}$$ is the only solution.

$$x^{\frac{1}{2}} + 3x^{\frac{1}{4}} - 18 = 0$$

\begin{align*} \left(x^{\frac{1}{4}} + 6\right)\left(x^{\frac{1}{4}} - 3\right) &= 0 \\ x^{\frac{1}{4}} + 6 = 0 & \text{ or } x^{\frac{1}{4}} - 3 = 0 \\ x^{\frac{1}{4}} = -6 & \text{ or } x^{\frac{1}{4}} = 3 \\ \left(x^{\frac{1}{4}}\right)^{4} = (-6)^{4} & \text{ or } \left(x^{\frac{1}{4}}\right)^{4} = (3)^{4} \\ x & = \text{1 296} \text{ or } 81 \end{align*}

We check both answers and find that $$x = 81$$ is the only solution.

$$\dfrac{16^x - 1}{4^2x + 1} = 3$$
\begin{align*} \frac{16^x - 1}{4^2x + 1} &= 3 \\ \frac{(4^{2x} - 1)(4^{2x} + 1)}{4^{2x} + 1} &= 3 \\ 4^{2x} - 1 &= 3 \\ 4^{2x} &= 4^1 \\ \therefore 2x &= 1 \\ x &= \frac{1}{2} \end{align*}
$$(2^x - 8)(3^x - 9) = 0$$
\begin{align*} (2^x - 8)(3^x - 9) &= 0 \\ (2^x - 2^3)(3^x - 3^2) &= 0 \\ \therefore 2^x - 2^3 = 0 &\text{ or } 3^x - 3^2 = 0 \\ \therefore x = 3 &\text{ or } x = 2 \end{align*}
$$(6^x - 36)(16 - 4^x) = 0$$
\begin{align*} (6^x - 36)(16 - 4^x) &= 0 \\ (6^x - 6^2)(4^2 - 4^x) &= 0 \\ \therefore 6^x - 6^2 = 0 &\text{ or } 4^2 - 4^x = 0 \\ \therefore x & = 2 \end{align*}
$$5.2^{x^2+1} = 20$$
\begin{align*} 5.2^{x^2+1} &= 20 \\ 2^{x^2+1} &= 4 \\ 2^{x^2+1} &= 2^2 \\ \therefore x^2+1 &= 2 \\ x^2 -1 &= 0 \\ (x+1)(x-1) & = 0 \\ \\ \therefore x = 1 &\text{ or } x = - 1 \end{align*}
$$27^{x-2} = 9^{2x+1}$$
\begin{align*} 27^{x-2} &= 9^{2x+1} \\ (3^3)^{x-2} &= (3^2)^{2x+1} \\ 3^{3x-6} &= 3^{4x+2} \\ \therefore 3x-6 &= 4x+2 \\ x &= -8 \end{align*}
$$\dfrac{8^x -1}{2^x - 1} = 7$$
\begin{align*} \frac{8^x -1}{2^x - 1} &= 7 \\ \frac{(2^3)^x -1}{2^x - 1} &= 7 \\ \frac{(2^x)^3 -1}{2^x - 1} &= 7 \\ \frac{(2^x -1)((2^x)^2 + 2^x + 1)}{(2^x - 1)} &= 7 \\ (2^{2x} + 2^x + 1) &= 7 \\ 2^{2x} + 2^x - 6 & = 0\\ \left(2^x + 3\right)\left(2^x - 2\right) &= 0 \\ \therefore 2^x + 3 = 0 & \text{ or } 2^x - 2 = 0 \\ 2^{x} \ne -3 & \text{ or } 2^{x} - 2 = 0 \\ & 2^{x} = 2 \\ & x = 1 \end{align*}
$$\dfrac{35^x}{5^x} = \dfrac{1}{7}$$
\begin{align*} \frac{35^x}{5^x} &= \frac{1}{7} \\ \frac{7^x5^x}{5^x} &= \frac{1}{7} \\ 7^x &= 7^{-1}\\ \therefore x &= -1 \end{align*}
$$\dfrac{a^{3x}\cdot a^{\frac{1}{x}}}{a^{-4}} = 1$$
\begin{align*} \frac{a^{3x}.a^{\frac{1}{x}}}{a^{-4}} &= 1\\ a^{3x+\frac{1}{x} + 4} &= a^0 \\ \therefore 3x+\frac{1}{x} + 4 &= 0 \\ 3x^2 + 1 + 4x &= 0 \\ (3x+1)(x+1) &= 0 \\ \therefore x = -\frac{1}{3} &\text{ or } x = -1 \end{align*}
$$2x^{\frac{1}{2}} + 1 = -x$$
\begin{align*} 2x^{\frac{1}{2}} + 1 &= -x\\ x + 2x^{\frac{1}{2}} + 1 &= 0 \\ (x^{\frac{1}{2}})^2 + 2x^{\frac{1}{2}} + 1^2 & = 0 \\ (x^{\frac{1}{2}} + 1)^2 &= 0 \\ x^{\frac{1}{2}} &= -1 \\ x &= 1 \end{align*}

However $$2(1)^{\frac{1}{2}} + 1 = 2 \neq -(1) ~ \therefore$$ no solution exists

Use trial and error to find the value of $$x$$ correct to 2 decimal places

$$4^x = 44$$
\begin{align*} 4^2 = 16 &\text{ and } 4^3 = 64 \\ \text{so } 2 < &x < 3 \\ \text{Test} \\ 4^{\text{2,5}} &= 32 \\ 4^{\text{2,75}} &= \text{45,255} \\ 4^{\text{2,70}} &= \text{42,224} \\ 4^{\text{2,73}} &= \text{44,017} \\ 4^{\text{2,725}} &= \text{43,713} \\ \therefore &x \approx \text{2,73} \end{align*}

Use trial and error to find the value of $$x$$ correct to 2 decimal places

$$3^x = 30$$
\begin{align*} 3^3 = 27 &\text{ and } 3^4 = 81 \\ \text{so } 3 < &x < 4 \\ \text{Test} \\ 3^{\text{3,1}} &= \text{30,014} \\ 3^{\text{3,05}} &= \text{28,525} \\ 3^{\text{3,08}} &= \text{29,480} \\ 3^{\text{3,09}} &= \text{29,806} \\ 3^{\text{3,095}} &= \text{29,970} \\ 3^{\text{3,096}} &= \text{30,003} \\ \therefore &x \approx \text{3,10} \end{align*}

Explain why the following statements are false:

$$\dfrac{1}{a^{-1} + b^{-1}}= a + b$$

The sum of two powers of the same degree is not the power of the sum of the bases

$a + b = \frac{1}{(a + b)^{-1}} \neq \frac{1}{a^{-1} + b^{-1}}$
$$(a + b)^2 = a^2 + b^2$$

The sum of two powers of the same degree is not the power of the sum of the bases

$(a + b)^2 = a^2 + 2ab + b^2 \neq a^2 + b^2$
$$\left(\frac{1}{a^2}\right)^{\frac{1}{3}} = a^{\frac{2}{3}}$$

A negative sign is missing, when a power is moved from the denominator to the numerator, the sign of the exponent changes.

From the question we must note that $$a \neq 0$$

\begin{align*} \left(\frac{1}{a^2}\right)^{\frac{1}{3}} & = \left(a^{-2}\right)^{\frac{1}{3}} \\ &= a^{-\frac{2}{3}} \end{align*}
$$2.3^x = 6^x$$

We cannot multiply bases unless they are raised to the same power

$6^x = (2 \times 3)^x = 2^x.3^x \neq 2.3^x$
$$x^{-\frac{1}{2}} = \dfrac{1}{-x^{\frac{1}{2}}}$$

The sign of a base is not changed when an exponent is moved from the denominator to the numerator in a fraction

$x^{-\frac{1}{2}} = \frac{1}{x^{\frac{1}{2}}} \neq \frac{1}{-x^{\frac{1}{2}}}$
$$(3x^4y^2)^3 = 3x^{12}y^6$$

The power of a product is the product of all the bases raised to the same power

\begin{align*} (3x^4y^2)^3 &= (3)^3(x^4)^3(y^2)^3 \\ &= 27x^{12}y^6 \neq 3x^{12}y^6 \end{align*}

If $$2^{2013}.5^{2015}$$ is written out in full how many digits will there be?

\begin{align*} 2^{2013}.5^{2015} &= 2^{2013}.5^{2013 + 2} \\ &= 2^{2013}.5^{2013}.5^2 \\ &= 25(2^{2013}.5^{2013}) \\ &= 25(10^{2013}) \\ &= 25\times 10^{2013} \end{align*}

$$10^{2013}$$ has $$\text{2 014}$$ digits therefore $$25 \times 10^{2013}$$ $$\text{2 015}$$ digits.

Prove that $$\dfrac{2^{n+1} + 2^n}{2^n - 2^{n-1}} = \dfrac{3^{n+1} + 3^n}{3^n - 3^{n-1}}$$

\begin{align*} \frac{2^{n+1} + 2^n}{2^n - 2^{n-1}} &= \frac{3^{n+1} + 3^n}{3^n - 3^{n-1}} \\ \\ \text{R.H.S} &= \frac{3^{n+1} + 3^n}{3^n - 3^{n-1}} \\ &= \frac{3^n(3^1 + 3^0)}{3^n(3^0 - 3^{-1})} \\ &= \frac{4}{1-\frac{1}{3}} \\ &= \frac{4}{\frac{2}{3}} \\ &= \frac{12}{2} \\ &= 6 \\ \\ \text{L.H.S} &= \frac{2^{n+1} + 2^n}{2^n - 2^{n-1}} \\ &= \frac{2^n(2^1 + 2^0)}{2^n(2^0 - 2^{-1})} \\ &= \frac{3}{1-\frac{1}{2}} \\ &= \frac{3}{\frac{1}{2}} \\ &= 6 \\ \\ \therefore \text{R.H.S} &= \text{L.H.S} \end{align*}