DAV Class 8 Maths Chapter 3 Worksheet 2
Exponents and Radicals Worksheet 2
1. Simplify
(i) \( x^{\frac{1}{2}} \times x^{\frac{5}{2}} \)
Solution
\[ \begin{align*} & = x^{\frac{1}{2} + \frac{5}{2}} \quad \color{magenta} \Big( a^m \times a^n = a^{m + n} \Big) \\ & = x^{\frac{1+5}{2}} \\ & = x^{\frac{\cancel6^3}{\cancel2_1}} \\ & = x^{3} \end{align*} \]Answer \( \color{red} x^{3} \)
(ii) \( \dfrac{x^{\frac{6}{5}}}{x^{\frac{1}{5}}} \)
Solution
\[ \begin{align*} & = x^{\frac{6}{5} - \frac{1}{5}} \quad \color{magenta}\Bigl(a^m \div a^n = a^{m - n}\Bigr) \\ & = x^{\frac{6 - 1}{5}} \\ & = x^{\frac{5}{5}} \\ & = x^{1} \end{align*} \]Answer \( \color{red}x \)
(iii) \( (x^7)^0 \)
Solution
\[ \begin{align*} & = x^{7 \times 0} \quad \color{magenta}\Bigl[(a^m)^n = a^{mn}\Bigr] \\ & = x^{0} \\ & = 1 \quad \color{magenta}(a^0 = 1) \end{align*} \]Answer \( \color{red}1 \)
(iv) \( 5\,x^{\frac{5}{6}} \times 6\,x^{\frac{1}{6}} \)
Solution
\[ \begin{align*} & = 5 \times 6 \times x^{\frac{5}{6} + \frac{1}{6}} \quad \color{magenta}\Bigl(a^m \times a^n = \color{magenta}a^{m + n}\Bigr) \\ & = 30 \times x^{\frac{5 + 1}{6}} \\ & = 30 \times x^{\frac{\cancel6^1}{\cancel6_1}} \\ & = 30 \times x^{1} \\ & = 30 x \end{align*} \]Answer \( \color{red}30x \)
(v) \( x^{-\frac{7}{2}} \times 2\,x^{-\frac{1}{2}} \)
Solution
\[ \begin{align*} & = 2 \times x^{-\frac{7}{2} + \bigl(-\frac{1}{2}\bigr)} \quad \color{magenta}\Bigl(a^m \times a^n = \color{magenta}a^{m + n}\Bigr) \\ \\ & = 2 \times x^{-\frac{7}{2} - \frac{1}{2}} \\ \\ & = 2 \times x^{\frac{-7 - 1}{2}} \\ \\ & = 2 \times x^{-\frac{\cancel8^4}{\cancel2_1}} \\ \\ & = 2 \times x^{-4} \\ \\ & = 2 \times \frac{1}{x^{4}} \quad \color{magenta}\Bigl(a^{-m} = \frac{1}{a^m}\Bigr) \\ \\ & = \frac{2}{x^4} \end{align*} \]Answer \( \color{red}\displaystyle \frac{2}{x^4} \)
2. Find the value of:
(i) \( (512)^{\frac{-2}{9}} \)
Solution
\[ \begin{align*} \color{magenta} Method \ - \ 1 \end{align*} \] \[ \begin{array}{c|c} 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array} \] \[ \begin{align*} 512 & = 2^9 \\ \\ & = (512)^{\frac{-2}{9}} \\ \\ & = (2^9)^{\frac{-2}{9}} \\ \\ & = (2)^{\cancel9^1 \times \frac{-2}{\cancel9_1}} \\ \\ & = 2^{-2} \\ \\ & = \frac{1}{2^{2}} \\ \\ & = \frac{1}{4} \\ \\ \end{align*} \] \[ \begin{align*} & \color{magenta} Method \ - \ 2 \\ \\ & = (512)^{\frac{-2}{9}} \\ \\ & = (8^3)^{\frac{-2}{9}} \\ \\ & = (8)^{\cancel3^1 \times \frac{-2}{\cancel9_3}} \\ \\ & = (8)^{\frac{-2}{3}} \\ \\ & = (2^3)^{\frac{-2}{3}} \\ \\ & = (2)^{\cancel3^1 \times \frac{-2}{\cancel3_1}} \\ \\ & = 2^{-2} \\ \\ & = \frac{1}{2^{2}} \\ \\ & = \frac{1}{4} \end{align*} \]Answer \( (512)^{\frac{-2}{9}} = \color{red} \dfrac{1}{4} \)
(ii) \( \Big[(216)^{\frac{2}{3}}\Big]^{\frac{1}{2}} \)
Solution
\[ \begin{align*} & = \Big[(216)^{\frac{2}{3}}\Big]^{\frac{1}{2}}\\ \\ & = \Big[(6^3)^{\frac{2}{3}}\Big]^{\frac{1}{2}}\\ \\ & = \Big[(6)^{\cancel3^1 \times \frac{2}{\cancel3_1}}\Big]^{\frac{1}{2}}\\ \\ & = \big[(6)^2\big]^{\frac{1}{2}}\\ \\ & = (6)^{\cancel2^1 \times \frac{1}{\cancel2_1}}\\ \\ & = 6 \end{align*} \]Answer \( \Big[(216)^{\frac{2}{3}}\Big]^{\frac{1}{2}} = \color{red} 6 \)
(iii) \( 10 \div 8^{\frac{-1}{3}} \)
Solution
\[ \begin{align*} & = 10 \div 8^{\frac{-1}{3}} \\ \\ & = 10 \div (2^3)^{\frac{-1}{3}} \\ \\ & = 10 \div (2)^{\cancel3^1 \times \frac{-1}{\cancel3_1}} \\ \\ & = 10 \div (2)^{-1} \\ \\ & = 10 \div \frac{1}{2} \\ \\ & = 10 \times 2 \\ \\ & = 20 \end{align*} \]Answer \( 10 \div 8^{\frac{-1}{3}} = \color{red} 20 \)
(iv) \( (16)^{\frac{3}{4}} \)
Solution
\[ \begin{align*} & = (16)^{\frac{3}{4}} \\ \\ & = (2^4)^{\frac{3}{4}} \\ \\ & = (2)^{\cancel4^1 \times \frac{3}{\cancel4_1}} \\ \\ & = 2^3 \\ \\ & = 8 \end{align*} \]Answer \( (16)^{\frac{3}{4}} = \color{red}8 \)
(v) \( 27^{\frac{1}{3}} \times 16^{-\frac{1}{4}} \)
Solution
\[ \begin{align*} & = 27^{\frac{1}{3}} \times 16^{-\frac{1}{4}} \\ \\ & = (3^3)^{\frac{1}{3}} \times (2^4)^{-\frac{1}{4}} \\ \\ & = (3)^{\cancel3^1 \times \frac{1}{\cancel3_1}} \times (2)^{\cancel4^1 \times \frac{-1}{\cancel4_1}} \\ \\ & = 3^1 \times 2^{-1} \\ \\ & = 3 \times \frac{1}{2} \\ \\ & = \frac{3}{2} \end{align*} \]Answer \( 27^{\frac{1}{3}}\times16^{-\frac{1}{4}} = \color{red}\dfrac{3}{2}\)
(vi) \( \dfrac{1}{\bigl[(3^4)^{\frac{1}{2}}\bigr]^{-2}} \)
Solution
\[ \begin{align*} & = \frac{1}{\bigl[(3^4)^{\tfrac{1}{2}}\bigr]^{-2}} \\ \\ & = \bigl[(3^4)^{\tfrac{1}{2}}\bigr]^2 \\ \\ & = \bigl[(3)^{\cancel4^2 \times \tfrac{1}{\cancel2_1}}\bigr]^2 \\ \\ & = \bigl[3^2\bigr]^2 \\ \\ & = 3^4 \\ \\ & = 81 \end{align*} \]Answer \( \dfrac{1}{\bigl[(3^4)^{\frac{1}{2}}\bigr]^{-2}} = \color{red}81\)
(vii) \( \dfrac{27^{\frac{-2}{3}}\times81^{\frac{5}{4}}}{\Biggl(\dfrac{1}{3}\Biggr)^{-3}} \)
Solution
\[ \begin{align*} & = \frac{27^{\frac{-2}{3}}\times81^{\frac{5}{4}}}{\Biggl(\dfrac{1}{3}\Biggr)^{-3}} \\ \\ & = \frac{(3^3)^{\frac{-2}{3}}\times(3^4)^{\frac{5}{4}}}{3^3} \\ \\ & = \frac{(3)^{\cancel3^1 \times \frac{-2}{\cancel3_1}}\times(3)^{\cancel4^1 \times \frac{5}{\cancel4^1}}}{3^3} \\ \\ & = \frac{3^{-2}\times3^5}{3^3} \\ \\ & = \frac{3^{-2+5}}{3^3} \\ \\ & = \frac{3^3}{3^3} \\ \\ & = 3^{3-3} \\ \\ & = 3^0 \\ \\ & = 1 \end{align*} \]Answer \( \dfrac{27^{\frac{-2}{3}}\times81^{\frac{5}{4}}}{\Biggl(\dfrac{1}{3}\Biggr)^{-3}} = \color{red}1\)
(viii) \( 64^{\frac{1}{2}}\bigl(64^{\frac{1}{2}}+1\bigr) \)
Solution
\[ \begin{align*} & = 64^{\frac{1}{2}} \times \bigl(64^{\frac{1}{2}}+1\bigr) \\ \\ & = (8^2)^{\frac{1}{2}} \times \bigl[(8^2)^{\frac{1}{2}}+1\bigr] \\ \\ & = (8)^{\cancel2^1 \times \frac{1}{\cancel2_1}} \times \bigl[(8)^{\cancel2^1 \times \frac{1}{\cancel2_1}}+1\bigr] \\ \\ & = 8\,(8+1) \\ \\ & = 8\times9 \\ \\ & = 72 \end{align*} \]Answer \( 64^{\frac{1}{2}}(64^{\frac{1}{2}}+1) = \color{red}72\)
(ix) \( \dfrac{36^{\frac{7}{2}}-36^{\frac{9}{2}}}{36^{\frac{5}{2}}} \)
Solution
\[ \begin{align*} & = \frac{36^{\frac{7}{2}}-36^{\frac{9}{2}}}{36^{\frac{5}{2}}} \\ \\ & = \frac{(6^2)^{\frac{7}{2}}-(6^2)^{\frac{9}{2}}}{(6^2)^{\frac{5}{2}}} \\ \\ & = \frac{(6)^{\cancel2^1 \times \frac{7}{\cancel2_1}}-(6)^{\cancel2^1 \times \frac{9}{\cancel2_1}}}{(6)^{\cancel2^1 \times \frac{5}{\cancel2_1}}} \\ \\ & = \frac{6^7 -6^9}{6^5} \\ \\ & = \frac{6^7\,(1 -6^2)}{6^5} \\ \\ & = {6^{7-5}\,(1 - 36)} \\ \\ & = {6^{2}\,(-35)} \\ \\ & = 36\,(-35) \\ \\ & = -1260 \end{align*} \]Answer \( \dfrac{36^{\frac{7}{2}}-36^{\frac{9}{2}}}{36^{\frac{5}{2}}} = \color{red}-1260\)
(x) \( 4\times81^{\frac{-1}{2}}\bigl(81^{\frac{1}{2}}+81^{\frac{3}{2}}\bigr) \)
Solution
\[ \begin{align*} & = 4 \times 81^{\frac{-1}{2}} \, \bigl(81^{\frac{1}{2}}+81^{\frac{3}{2}}\bigr) \\ \\ & = 4 \times (9^2)^{\frac{-1}{2}} \, \Bigl[(9^2)^{\frac{1}{2}}+(9^2)^{\frac{3}{2}}\Bigr] \\ \\ & = 4 \times (9)^{\cancel2^1 \times \frac{-1}{\cancel2_1}} \, \Bigl[(9)^{\cancel2^1 \times \frac{1}{\cancel2_1}}+(9)^{\cancel2^1 \times \frac{3}{\cancel2_1}}\Bigr] \\ \\ & = 4 \times 9^{-1} \, \Bigl[9 ^1 + 9^3 \Bigr] \\ \\ & = 4 \times \Bigl[9^{-1} \times 9^1 + 9^{-1} \times 9^3 \Bigr] \\ \\ & = 4 \times \Bigl[9^{-1 + 1} + 9^{-1 + 3}\Bigr] \\ \\ & = 4 \times \Bigl[9^0 + 9^2\Bigr] \\ \\ & = 4 \times \Bigl[1 + 81 \Bigr] \\ \\ & = 4 \times 82 \\ \\ & = 328 \end{align*} \]Answer \( 4\times81^{-\frac{1}{2}}(81^{\frac{1}{2}}+81^{\frac{3}{2}}) = \color{red}328\)
3. Evaluate:
(i) \( (0.04)^{\frac{3}{2}} \)
Solution
\[ \begin{align*} & = (0.04)^{\frac{3}{2}} \\ & = \Bigg(\frac{4}{100}\Bigg)^{\tfrac{3}{2}} \\ & = \Bigg(\frac{2^2}{10^2}\Bigg)^{\tfrac{3}{2}} \\ & = \Bigg[\Bigg(\frac{2}{10}\Bigg)^2 \ \Bigg]^{\tfrac{3}{2}} \\ & = \Bigg(\frac{2}{10}\Bigg)^{\cancel2^1 \times \tfrac{3}{\cancel2_1}} \\ & = \Bigg(\frac{2}{10}\Bigg)^3 \\ & = \Bigg(\frac{8}{1000}\Bigg) \\ \\ & = 0.008 \end{align*} \]Answer \( (0.04)^{\frac{3}{2}} = \color{red} 0.008 \)
(ii) \( (6.25)^{\frac{3}{2}} \)
Solution
\[ \begin{align*} & = (6.25)^{\frac{3}{2}} \\ \\ & = \Bigg(\frac{625}{100}\Bigg)^{\tfrac{3}{2}} \\ & = \Bigg(\frac{25^2}{10^2}\Bigg)^{\tfrac{3}{2}} \\ & = \Bigg[\Bigg(\frac{25}{10}\Bigg)^2 \ \Bigg]^{\tfrac{3}{2}} \\ & = \Bigg(\frac{25}{10}\Bigg)^{\cancel2^1 \times \tfrac{3}{\cancel2_1}} \\ & = \Bigg(\frac{25}{10}\Bigg)^3 \\ \\ & = \frac{15625}{1000} \\ \\ & = 15.625 \end{align*} \]Answer \( (6.25)^{\frac{3}{2}} = \color{red}15.625 \)
(iii) \( (0.03125)^{\frac{-2}{5}} \)
Solution
\[ \begin{align*} & = (0.03125)^{\frac{-2}{5}} \\ \\ & = \Bigg(\frac{3125}{100000}\Bigg)^{\tfrac{-2}{5}} \\ \\ & = \Bigg(\frac{100000}{3125}\Bigg)^{\tfrac{2}{5}} \\ \\ & = \Bigg[\Bigg(\frac{10^5}{5^5}\Bigg)\Bigg]^{\tfrac{2}{5}} \\ \\ & = \Bigg[\Bigg(\frac{10}{5}\Bigg)^5 \, \Bigg]^{\tfrac{2}{5}} \\ \\ & = \Bigg(\frac{10}{5}\Bigg)^{\cancel5^1 \times \tfrac{2}{\cancel5_1}} \\ \\ & = \Bigg(\frac{\cancel{10}^2}{\cancel5_1}\Bigg)^2 \\ \\ & = 2^2 \\ \\ & = 4 \end{align*} \]Answer \( (0.03125)^{-\frac{2}{5}} = \color{red}4 \)
(iv) \( (0.008)^{\frac{2}{3}} \)
Solution
\[ \begin{align*} & = (0.008)^{\frac{2}{3}} \\ \\ & = \Bigg(\frac{8}{1000}\Bigg)^{\tfrac{2}{3}} \\ \\ & = \Bigg(\frac{2^3}{10^3}\Bigg)^{\tfrac{2}{3}} \\ \\ & = \Bigg[\Bigg(\frac{2}{10}\Bigg)^3\,\Bigg]^{\tfrac{2}{3}} \\ \\ & = \Bigg(\frac{2}{10}\Bigg)^{\cancel3^1 \times \tfrac{2}{\cancel3_1}} \\ \\ & = \Bigg(\frac{2}{10}\Bigg)^{2} \\ \\ & = \frac{4}{100} \\ \\ & = 0.04 \end{align*} \]Answer \( (0.008)^{\frac{2}{3}} = \color{red}0.04 \)
4. Evaluate:
(i) \( (6^2 + 8^2)^{\frac{1}{2}} \)
Solution
\[ \begin{align*} & = (6^2 + 8^2)^{\frac{1}{2}} \\ & = (36 + 64)^{\frac{1}{2}} \\ & = (100)^{\frac{1}{2}} \\ & = (10^2)^{\frac{1}{2}} \\ & = (10)^{2 \times \frac{1}{2}} \\ & = 10 \end{align*} \]Answer \( (6^2 + 8^2)^{\frac{1}{2}} = \color{red}10 \)
(ii) \( \Big[5 \big(8^{\frac{1}{3}} + 27^{\frac{1}{3}}\big)^3 \Big]^\frac{1}{4} \)
Solution
\[ \begin{align*} & = \Big[ 5 \big(8^{\frac{1}{3}} + 27^{\frac{1}{3}}\big)^3 \Big]^\frac{1}{4} \\ & = \Big[ 5 \big(2^{\cancel3^1 \times \frac{1}{\cancel3_1}} + 3^{\cancel3^1 \times\frac{1}{\cancel3_1}}\big)^3 \Big]^\frac{1}{4} \\ & = \Big[ 5 \big(2 + 3\big)^3 \Big]^\frac{1}{4} \\ & = \Big[ 5 \big(5\big)^3 \Big]^\frac{1}{4} \\ & = \Big[ 5 \times 5^3 \Big]^\frac{1}{4} \\ & = \Big[ 5^4 \Big]^\frac{1}{4} \\ & = 5^{\cancel4^1 \times \frac{1}{\cancel4_1}} \\ & = 5 \end{align*} \]Answer \( \Big[5 \big(8^{\frac{1}{3}} + 27^{\frac{1}{3}}\big)^3 \Big]^\frac{1}{4} = \color{red} 5 \)
(iii) \( (17^2 - 8^2)^{\tfrac{1}{2}} \)
Solution
\[ \begin{align*} & = (17^2 - 8^2)^{\tfrac{1}{2}} \\ & = (289 - 64)^{\tfrac{1}{2}} \\ & = (225)^{\tfrac{1}{2}} \\ & = (15^2)^{\tfrac{1}{2}} \\ & = (15)^{\cancel2^1 \times \tfrac{1}{\cancel2_1}} \\ \\ & = 15 \end{align*} \]Answer \( (17^2 - 8^2)^{\tfrac{1}{2}} = \color{red}15 \)
(iv) \( (1^3 + 2^3 + 3^3)^{\frac{-5}{2}} \)
Solution
\[ \begin{align*} & = (1^3 + 2^3 + 3^3)^{\frac{-5}{2}} \\ \\ & = (1 + 8 + 27)^{\frac{-5}{2}} \\ \\ & = (36)^{\frac{-5}{2}} \\ \\ & = (6^2)^{\frac{-5}{2}} \\ \\ & = 6^{\cancel2^1 \times \frac{-5}{\cancel2_1}} \\ \\ & = 6^{-5} \\ \\ & = \frac{1}{6^5} \\ \\ & = \frac{1}{7776} \end{align*} \]Answer \( (1^3 + 2^3 + 3^3)^{\frac{-5}{2}} = \color{red}\dfrac{1}{7776} \)
5. Simplify and express the answers with positive indices:
(i) \( 2x^{\frac{1}{6}} \times 2x^{\frac{-7}{6}} \)
Solution
\[ \begin{align*} & = 2x^{\frac{1}{6}} \times 2x^{\frac{-7}{6}} \\ & = 2 \times 2 \times x^{\frac{1}{6}} \times x^{\frac{-7}{6}} \\ & = 4 \times x^{\frac{1}{6} - \frac{7}{6} } \\ & = 4 \times x^{\frac{1-7}{6} } \\ & = 4 \times x^{\frac{-\cancel6^1}{\cancel6_1} } \\ & = 4 \times x^{-1} \\ & = 4 \times \frac{1}{x} \\ & = \frac{4}{x} \end{align*} \]Answer \( 2x^{\frac{1}{6}} \times 2x^{\frac{-7}{6}} = \color{red} \dfrac{4}{x} \)
(ii) \( \left[\sqrt[4]{\left(\dfrac{1}{x}\right)^{-12}}\right]^{\tfrac{-2}{3}} \)
Solution
\[ \begin{align*} & = \left[\sqrt[4]{\left(\frac{1}{x}\right)^{-12}}\right]^{\tfrac{-2}{3}} \\ \\ & = \left[\left(\frac{1}{x}\right)^{\tfrac{-\cancel{12}^3}{\cancel4_1}}\right]^{\tfrac{-2}{3}} \\ \\ & = \left[\left(\frac{1}{x}\right)^{-3}\right]^{\tfrac{-2}{3}} \\ \\ & = \left(\frac{1}{x}\right)^{-\cancel3^1 \times \tfrac{-2}{\cancel3_1}} \\ \\ & = \left(\frac{1}{x}\right)^{2} \\ \\ & = \frac{1}{x^2} \\ \\ \end{align*} \]Answer \( \left[\sqrt[4]{\left(\dfrac{1}{x}\right)^{-12}}\right]^{\tfrac{-2}{3}} = \color{red} \dfrac{1}{x^2} \)
(iii) \( a^{\frac{4}{7}} \div a^{\frac{10}{7}} \)
Solution
\[ \begin{align*} & = a^{\frac{4}{7}} \div a^{\frac{10}{7}} \\ \\ & = a^{\frac{4}{7} - \frac{10}{7}} \\ \\ & = a^{\frac{4-10}{7}} \\ \\ & = a^{\frac{4-10}{7}} \\ \\ & = a^{\frac{-6}{7}} \\ \\ & = \frac{1}{a^{\frac{6}{7}}} \\ \\ \end{align*} \]Answer \( a^{\frac{4}{7}} \div a^{\frac{10}{7}} = \color{red} \dfrac{1}{a^{\frac{6}{7}}} \)
6. Verify that: \( \Big[(729)^{\frac{-5}{3}} \Big]^\frac{-1}{2} = (729)^{\frac{-5}{3} \times \left(\frac{-1}{2}\right)} \)
Solution
\[ \begin{align*} LHS & = \Big[(729)^{\frac{-5}{3}} \Big]^\frac{-1}{2} \\ & = \Big[(3^6)^{\frac{-5}{3}} \Big]^\frac{-1}{2} \\ & = \Big[3^{\cancel6^2 \times \frac{-5}{\cancel3_1}} \Big]^\frac{-1}{2} \\ & = \Big[3^{-10} \Big]^\frac{-1}{2} \\ & = 3^{\cancel{-10}^{-5} \times \frac{-1}{\cancel2_1}} \\ & = 3^{5} \\ & = 243 \\ \\ RHS & = (729)^{\frac{-5}{3} \times \left(\frac{-1}{2}\right)} \\ & = (729)^{\frac{5}{6} } \\ & = (3^6)^{\frac{5}{6} } \\ & = (3)^{\cancel6^1 \times \frac{5}{\cancel6_1} } \\ & = 3^{5} \\ & = 243 \\ \\ LHS & = RHS \\ \Big[(729)^{\frac{-5}{3}} \Big]^\frac{-1}{2} & = (729)^{\frac{-5}{3} \times \left(\frac{-1}{2}\right)} \\ Hence & \ \ Verified \end{align*} \]7. Solve the given exponential equations:
(i) \( (\sqrt 6)^{x-2} = 1 \)
Solution
\[ \begin{align*} (\sqrt 6)^{x-2} &= 1 \\ (6^{\frac{1}{2}})^{x-2} &= 1 \\ 6^{\frac{x-2}{2}} &= 1 \\ 6^{\frac{x-2}{2}} &= 6^0 \\ \text{Bases are the same, } & \text{we equate powers.} \\ \frac{x-2}{2} & = 0 \\ x - 2 & = 0 \\ x & = 2 \end{align*} \]Answer \( x = \color{red} 2 \)
(ii) \( 3^{4x} = \dfrac{1}{81} \)
Solution
\[ \begin{align*} 3^{4x} &= \frac{1}{81} \\[6pt] 3^{4x} &= \frac{1}{3^4} \\[6pt] 3^{4x} &= 3^{-4} \\[6pt] \text{Bases are the same, } & \text{we equate powers.} \\ 4x &= -4 \\[6pt] x &= \frac{-4}{4} \\[6pt] x & = -1 \end{align*} \]Answer \( x = \color{red}-1 \)
(iii) \( (\sqrt{2})^{x} = 2^{8} \)
Solution
\[ \begin{align*} (\sqrt{2})^x &= 2^8 \\[6pt] (2^{\tfrac12})^x &= 2^8 \\[6pt] 2^{\tfrac{x}{2}} &= 2^8 \\[6pt] \text{Bases are the same, } & \text{we equate powers.} \\ \frac{x}{2} &= 8 \\[6pt] x &= 16 \end{align*} \]Answer \( x = \color{red}16 \)
(iv) \( 2^{2x+1} = 4^{2x-1} \)
Solution
\[ \begin{align*} 2^{2x+1} &= 4^{2x-1} \\[6pt] 2^{2x+1} &= (2^2)^{2x-1} \\[6pt] 2^{2x+1} &= 2^{2(2x-1)} \\[6pt] 2^{2x+1} &= 2^{4x-2} \\[6pt] \text{Bases are the same, } & \text{we equate powers.} \\ 2x+1 &= 4x-2 \\[6pt] 1 + 2 &= 4x - 2x \\[6pt] 3 &= 2x \\[6pt] x &= \frac{3}{2} \end{align*} \]Answer \( x = \color{red}\dfrac{3}{2} \)