DAV Class 8 Maths Chapter 3 Brain Teasers
Exponents and Radicals Brain Teasers
1. A. Tick (✓) the correct option.
a.
Value of \( {[(625)^{\frac{-1}{2}}]^2 } \)
\(
\begin{aligned}
(i) \, & \, \frac{1}{625} \\[5pt]
(ii) \, & \, -25 \\[5pt]
(iii) \, & \, -625 \\[5pt]
(iv) \, & \, 25 \\[5pt]
\end{aligned}
\)
Solution
\[ \begin{align*} & = {[(625)^{\frac{-1}{2}}]^2 } \\[5pt] & = {(625)^{\frac{-1}{\cancel2_1} \times \cancel2^1} } \\[5pt] & = {(625)^{-1}} \\[5pt] & = \dfrac{1}{625} \end{align*} \]
Answer \( \color{orange}(i)\ \color{red} \dfrac{1}{625} \)
b.
If \( 5^x = 1 \), value of \( x \) is
\(
\begin{aligned}
(i) \, & \, 5 \\[5pt]
(ii) \, & \, \frac{1}{25} \\[5pt]
(iii) \, & \, 0 \\[5pt]
(iv) \, & \, \frac{1}{5} \\[5pt]
\end{aligned}
\)
Solution
\[ \begin{align*} 5^x &= 1 \\ 5^x &= 5^0 \\ \implies x &= 0 \\ \end{align*} \]
Answer \( \color{orange}(iii)\ \color{red} 0 \)
c.
Value of \( (27^{\frac{1}{3}} + 64^{\frac{1}{3}})^2 \) is
\(
\begin{aligned}
(i) \, & \, 7^{\frac{1}{3}} \\[5pt]
(ii) \, & \, 49 \\[5pt]
(iii) \, & \, 7^{\frac{1}{2}} \\[5pt]
(iv) \, & \, \frac{1}{7} \\[5pt]
\end{aligned}
\)
Solution
\[ \begin{align*} & = (27^{\frac{1}{3}} + 64^{\frac{1}{3}})^2 \\[5pt] & = (3^{\cancel3^1 \times \frac{1}{\cancel3_1}} + 4^{\cancel3^1 \times \frac{1}{\cancel3_1}})^2 \\[5pt] & = (3 + 4)^2 \\[5pt] & = (7)^2 \\[5pt] & = 49 \end{align*} \]
Answer \( \color{orange}(ii)\ \color{red} 49 \)
d.
For any two non-zero rational numbers a and b, \( a^4 \div b^4 \) is equal to
\(
\begin{aligned}
(a) \, & \, (a \div b)^1 \\
(b) \, & \, (a \div b)^0 \\
(c) \, & \, (a \div b)^4 \\
(d) \, & \, (a \div b)^8 \\
\end{aligned}
\)
Solution
\[ \begin{align*} &= a^4 \div b^4 \\ &= (a \div b)^4 \\ \end{align*} \]
Answer \( \color{orange}(iii)\ \color{red} (a \div b)^4 \)
e.
\(4.25 \times 10^{-7} \)
\(
\begin{aligned}
(a) \, & \, 0.425000000 \\
(b) \, & \, 4250000 \\
(c) \, & \, 0.000000425 \\
(d) \, & \, 0.00000425 \\
\end{aligned}
\)
Solution
\[ \begin{align*} & = 4.25 \times 10^{-7} \\[5pt] & = 4.25 \times \frac{1}{10^7} \\[5pt] & = \frac{4.25}{10000000} \\[5pt] & = 0.000000425 \end{align*} \]
Answer \( \color{orange}(iii)\ \color{red} 0.000000425 \)
B. Answer the following questions.
(a) Express \( \left(\dfrac{13}{21}\right)^{\frac{2}{5}} \) as a radical.
Answer \( \color{red} \sqrt[5] {\left(\dfrac{13}{21}\right)^2} \)
(b) Find the value of \( (81)^{\frac{-3}{4}} \)
Solution
\[ \begin{align*} & = (81)^{\frac{-3}{4}} \\[5pt] & = (3)^{\cancel4^1 \times \frac{-3}{\cancel4_1}} \\[5pt] & = 3^{-3} \\[5pt] & = \frac{1}{3^3} \\[5pt] & = \frac{1}{27} \end{align*} \]
Answer \( \color{red} \dfrac{1}{27} \)
(c) Simplify \( 5 \times 16^{\frac{3}{4}} \)
Solution
\[ \begin{align*} & = 5 \times 16^{\frac{3}{4}} \\[5pt] & = 5 \times 2^{\cancel4^1 \times \frac{3}{\cancel4_1}} \\[5pt] & = 5 \times 2^3 \\[5pt] & = 5 \times 8 \\[5pt] & = 40 \end{align*} \]
Answer \( \color{red} 40 \)
(d) Evaluate \( (0.000064)^{\frac{5}{6}} \)
Solution
\[ \begin{align*} & = (0.000064)^{\frac{5}{6}} \\[5pt] & = \left(\frac{64}{1000000}\right)^{\frac{5}{6}} \\[5pt] & = \left(\frac{2^6}{10^6}\right)^{\frac{5}{6}} \\[5pt] & = \left(\frac{2}{10}\right)^{\cancel6^1 \times \frac{5}{\cancel6_1}} \\[5pt] & = \left(\frac{2}{10}\right)^5 \\[5pt] & = \frac{32}{100000} \\[5pt] & = 0.00032 \end{align*} \]
Answer \( \color{red} 0.00032 \)
(e) Solve for \( x \) if \( 3^{x-1} = \dfrac{1}{27} \)
Solution
\[ \begin{align*} 3^{x-1} &= \dfrac{1}{27} \\[5pt] 3^{x-1} &= \dfrac{1}{3^3} \\[5pt] 3^{x-1} &= 3^{-3} \\[5pt] \implies x-1 &= -3 \\[5pt] x &= -3+1 \\[5pt] x &= -2 \end{align*} \]
Answer \( x = \color{red} -2 \)
2. Simplify \( \dfrac{(64)^{\frac{-1}{6}} \times (216)^{\frac{-1}{3}} \times (81)^{\frac{1}{4}}}{(512)^{\frac{-1}{3}} \times (16)^{\frac{1}{4}} \times (9)^{\frac{-1}{2}}} \)
Solution
\[ \begin{aligned} & = \frac{(64)^{\frac{-1}{6}} \times (216)^{\frac{-1}{3}} \times (81)^{\frac{1}{4}}}{(512)^{\frac{-1}{3}} \times (16)^{\frac{1}{4}} \times (9)^{\frac{-1}{2}}} \\ \\ & = \frac{(2^6)^{\frac{-1}{6}} \times (6^3)^{\frac{-1}{3}} \times (3^4)^{\frac{1}{4}}}{(8^3)^{\frac{-1}{3}} \times (2^4)^{\frac{1}{4}} \times (3^2)^{\frac{-1}{2}}} \\ \\ & = \frac{(2)^{\cancel6 \times \frac{-1}{\cancel6}} \times (6)^{\cancel3 \times \frac{-1}{\cancel3}} \times (3)^{\cancel4 \times \frac{1}{\cancel4}}}{(8)^{\cancel3 \times \frac{-1}{\cancel3}} \times (2)^{\cancel4 \times \frac{1}{\cancel4}} \times (3)^{\cancel2 \times \frac{-1}{\cancel2}}} \\ \\ & = \frac{2^{-1} \times 6^{-1} \times 3^{1}}{8^{-1} \times 2^{1} \times 3^{-1}} \\ \\ & = \frac{8 \times 3 \times \cancel3^1}{2 \times 2 \times \cancel6_2} \\ \\ & = \frac{\cancel8^1 \times 3}{\cancel8_1} \\ \\ & = 3 \end{aligned} \]
Answer \( \color{red} 3 \)
3. Simplify and express the answer with positive exponent: \( \left[\sqrt[3]{x^4 \ y \ } \times \dfrac{1}{\sqrt[3]{x \ y^7 \ }}\right]^{-4} \)
Solution
\[ \begin{aligned} & = \left[\sqrt[3]{x^4 \ y \ } \times \frac{1}{\sqrt[3]{x \ y^7 \ }}\right]^{-4} \\ \\ & = \left[\frac{\sqrt[3]{x^4 \ y \ }}{\sqrt[3]{x \ y^7 \ }}\right]^{-4} \\ \\ & = \left[\frac{({x^4 \ y \ })^{\frac{1}{3}}}{({x \ y^7 \ })^{\frac{1}{3}}}\right]^{-4} \\ \\ & = \left[\left(\frac{{x^4 \ y \ }}{{x \ y^7 \ }}\right)^{\frac{1}{3}}\right]^{-4} \\ \\ & = \left[\left({x^{(4-1)} \ y^{(1-7)}}\right)^{\frac{1}{3}}\right]^{-4} \\ \\ & = \left[\left({x^{3} \ y^{-6}}\right)^{\frac{1}{3}}\right]^{-4} \\ \\ & = \left[{x^{\cancel3^1 \times \frac{1}{\cancel3_1}} \ y^{-\cancel{6}^2 \times \frac{1}{\cancel3_1} } \ } \right]^{-4} \\ \\ & = \left[{x^1 \ y^{-2} \ } \right]^{-4} \\ \\ & = {x^{-4} \ y^{8} \ }\\ \\ & = \frac{y^{8}}{x^4}\\ \\ \end{aligned} \]
Answer \( \color{red} \dfrac{y^{8}}{x^4} \)
4. Evaluate:
(i) \(3 \times 16^{\frac{3}{4}}\)
Solution
\[ \begin{aligned} & = 3 \times 16^{\frac{3}{4}} \\[5pt] &= 3 \times (2^4)^{\frac{3}{4}} \\[5pt] &= 3 \times 2^{\cancel4 \times \frac{3}{\cancel4}} \\[5pt] &= 3 \times 2^3 \\[5pt] &= 3 \times 8 \\[5pt] &= 24 \end{aligned} \]
Answer \( \color{red} 24 \)
(ii) \(2 \times 27^{\frac{-2}{3}}\)
Solution
\[ \begin{aligned} & = 2 \times 27^{\frac{-2}{3}} \\[5pt] & = 2 \times (3^3)^{\frac{-2}{3}} \\[5pt] & = 2 \times 3^{\cancel3 \times \bigl(\frac{-2}{\cancel3}\bigr)} \\[5pt] & = 2 \times 3^{-2} \\[5pt] & = 2 \times \frac{1}{3^2} \\[5pt] & = 2 \times \frac{1}{9} \\[5pt] & = \frac{2}{9} \end{aligned} \]
Answer \( \color{red} \dfrac{2}{9}\)
(iii) \(2 \times 9^{\frac{3}{2}} \times 9^{\frac{-1}{2}}\)
Solution
\[ \begin{aligned} & = 2 \times 9^{\frac{3}{2}} \times 9^{\frac{-1}{2}} \\[5pt] & = 2 \times 9^{\frac{3}{2} + \bigl(\frac{-1}{2}\bigr)} \\[5pt] & = 2 \times 9^{\frac{3}{2} - \frac{1}{2}} \\[5pt] & = 2 \times 9^{\frac{3-1}{2}} \\[5pt] & = 2 \times 9^{\frac{2}{2}} \\[5pt] & = 2 \times 9 \\[5pt] & = 18 \end{aligned} \]
Answer \( \color{red} 18\)
5. Find the value of \( [5^2 + 12^2]^{\frac{1}{2}} \)
Solution
\[ \begin{aligned} & = [5^2 + 12^2]^{\frac{1}{2}} \\[5pt] & = [25 + 144]^{\frac{1}{2}} \\[5pt] & = [169]^{\frac{1}{2}} \\[5pt] & = [13^2]^{\frac{1}{2}} \\[5pt] & = [13]^{\cancel2 \times \frac{1}{\cancel2}} \\[5pt] & = 13 \end{aligned} \]
Answer \( \color{red} 13 \)
6. Find the value of \( x \) if
(i) \(2^x + 2^x + 2^x = 192\)
Solution
\[ \begin{aligned} 2^x + 2^x + 2^x &= 192 \\[5pt] 3 \times 2^x &= 192 \\[5pt] 2^x &= \frac{192}{3} \\[5pt] 2^x & = 64 \\[5pt] 2^x &= 2^6 \\[5pt] \text{Bases are the same, } & \text{we equate powers.} \\ x & = 6 \end{aligned} \]
Answer \( x= \color{red} 6 \)
(ii) \(8^{255} = 32^x\)
Solution
\[ \begin{aligned} 8^{255} &= 32^x \\[5pt] (2^3)^{255} &= (2^5)^x \\[5pt] 2^{3\times 255} &= 2^{5x} \\[5pt] 2^{765} &= 2^{5x} \\[5pt] \text{Bases are the same, } & \text{we equate powers.} \\ 765 &= 5x \\[5pt] x &= \frac{765}{5} \\[5pt] x &= 153 \end{aligned} \]
Answer \( x= \color{red} 153 \)
(iii) \(2^{2x+2} = 4^{2x-1}\)
Solution
\[ \begin{aligned} 2^{2x+2} &= 4^{2x-1} \\[5pt] 2^{2x+2} &= (2^2)^{2x-1} \\[5pt] 2^{2x+2} &= 2^{2(2x-1)} \\[5pt] 2^{2x+2} &= 2^{4x-2} \\[5pt] \text{Bases are the same, } & \text{we equate powers.} \\ 2x+2 &= 4x-2 \\[5pt] 2 + 2 & = 4x - 2x \\[5pt] 4 &= 2x \\[5pt] \frac{4}{2} &= x \\[5pt] x & = 2 \end{aligned} \]
Answer 2
7. If \(4^x - 4^{x-1} = 24\), then find \(x\).
Solution
\[ \begin{align*} 4^x - 4^{x-1} & = 24 \\ \\ 4^x - \frac{4^x}{4} & = 24 \\ \\ 4^x \left( 1 - \frac{1}{4} \right) & = 24 \\ \\ 4^x \left( 1 - \frac{1}{4} \right) & = 24 \\ \\ 4^x \left(\frac{4 - 1}{4} \right) & = 24 \\ \\ 4^x \left(\frac{3}{4} \right) & = 24 \\ \\ 4^x & = \cancel{24}^8 \times \frac{4}{\cancel3_1} \\ \\ 4^x & = 32 \\[6pt] (2^2)^x & = 2^5 \\[6pt] 2^{2x} & = 2^5 \\[6pt] \text{Bases are the same, } & \text{we equate powers.} \\ 2x & = 5 \\[6pt] x & = \frac{5}{2} \\[6pt] \end{align*} \]
Answer \(x = \color{red} \dfrac{5}{2} \)