Unit 9: Indices (घाताङ्क) Formulae & Solutions
Solutions by Important Edu Notes Team
1. Laws of Indices (घाताङ्कका नियमहरू)
[Image of laws of exponents chart]| Name | Rule (नियम) | Example |
|---|---|---|
| Product Rule | $$a^m \times a^n = a^{m+n}$$ | $$2^3 \times 2^2 = 2^5$$ |
| Quotient Rule | $$\frac{a^m}{a^n} = a^{m-n}$$ | $$\frac{5^5}{5^2} = 5^3$$ |
| Power Rule | $$(a^m)^n = a^{mn}$$ | $$(3^2)^3 = 3^6$$ |
| Negative Index | $$a^{-m} = \frac{1}{a^m}$$ | $$2^{-3} = \frac{1}{8}$$ |
| Zero Index | $$a^0 = 1 \quad (a \neq 0)$$ | $$100^0 = 1$$ |
| Root Rule | $$\sqrt[n]{a^m} = a^{\frac{m}{n}}$$ | $$\sqrt{x} = x^{\frac{1}{2}}$$ |
Solving Exponential Equations:
- Base Equality: If $a^x = a^y$ and $a \neq 1$, then $x = y$.
- Power Equality: If $x^a = y^a$ and $a \neq 0$, then $x = y$.
2. Exercise 9.1 – Q1: Tables
Q1.a: Fill in the table for $y = 7^x$.
We calculate values of $7^x$ for each $x$:
$x = -3 \Rightarrow 7^{-3} = \frac{1}{343}$
$x = -2 \Rightarrow 7^{-2} = \frac{1}{49}$
$x = -1 \Rightarrow 7^{-1} = \frac{1}{7}$
$x = 0 \Rightarrow 7^0 = 1$
$x = 1 \Rightarrow 7^1 = 7$
$x = 2 \Rightarrow 7^2 = 49$
$x = 3 \Rightarrow 7^3 = 343$
$x = -2 \Rightarrow 7^{-2} = \frac{1}{49}$
$x = -1 \Rightarrow 7^{-1} = \frac{1}{7}$
$x = 0 \Rightarrow 7^0 = 1$
$x = 1 \Rightarrow 7^1 = 7$
$x = 2 \Rightarrow 7^2 = 49$
$x = 3 \Rightarrow 7^3 = 343$
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| $7^x$ | 1/343 | 1/49 | 1/7 | 1 | 7 | 49 | 343 |
Q1.b: Fill in the table for $y = 5^{-x}$.
We calculate values of $5^{-x}$:
$x = -3 \Rightarrow 5^{-(-3)} = 125$
$x = -2 \Rightarrow 5^2 = 25$
$x = -1 \Rightarrow 5^1 = 5$
$x = 0 \Rightarrow 5^0 = 1$
$x = 1 \Rightarrow 5^{-1} = 1/5$
$x = 2 \Rightarrow 5^{-2} = 1/25$
$x = 3 \Rightarrow 5^{-3} = 1/125$
$x = -2 \Rightarrow 5^2 = 25$
$x = -1 \Rightarrow 5^1 = 5$
$x = 0 \Rightarrow 5^0 = 1$
$x = 1 \Rightarrow 5^{-1} = 1/5$
$x = 2 \Rightarrow 5^{-2} = 1/25$
$x = 3 \Rightarrow 5^{-3} = 1/125$
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| $5^{-x}$ | 125 | 25 | 5 | 1 | 1/5 | 1/25 | 1/125 |
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3. Exercise 9.1 – Q2: Solve and Examine
Q2.a: Solve and examine: $3^x = 9$
$$3^x = 3^2$$
$$x = 2$$
$$x = 2$$
Check: $3^2 = 9$ (Correct)
x = 2
Q2.b: Solve and examine: $5^{x-1} = 25$
$$5^{x-1} = 5^2$$
$$x – 1 = 2$$
$$x = 3$$
$$x – 1 = 2$$
$$x = 3$$
Check: $5^{3-1} = 5^2 = 25$ (Correct)
x = 3
Q2.c: Solve and examine: $\frac{1}{5^{2x-4}} = 125$
$$5^{-(2x-4)} = 5^3$$
$$-2x + 4 = 3$$
$$-2x = -1$$
$$x = 1/2$$
$$-2x + 4 = 3$$
$$-2x = -1$$
$$x = 1/2$$
x = 1/2
Q2.d: Solve and examine: $4^{x-2} = 0.125$
$0.125 = 1/8 = 2^{-3}$ and $4 = 2^2$
$$(2^2)^{x-2} = 2^{-3}$$
$$2^{2x-4} = 2^{-3}$$
$$2x – 4 = -3$$
$$2x = 1 \Rightarrow x = 1/2$$
$$2^{2x-4} = 2^{-3}$$
$$2x – 4 = -3$$
$$2x = 1 \Rightarrow x = 1/2$$
x = 1/2
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Q2.e: Solve and examine: $(\frac{3}{5})^x = (1\frac{2}{3})^3$
$1\frac{2}{3} = \frac{5}{3}$
$$(\frac{3}{5})^x = (\frac{5}{3})^3$$
$$(\frac{3}{5})^x = (\frac{3}{5})^{-3}$$
$$x = -3$$
$$(\frac{3}{5})^x = (\frac{3}{5})^{-3}$$
$$x = -3$$
x = -3
Q2.f: Solve and examine: $2^x \times 3^{x+1} = 18$
$$2^x \times 3^x \times 3^1 = 18$$
$$(2 \times 3)^x \times 3 = 18$$
$$6^x = 6$$
$$x = 1$$
$$(2 \times 3)^x \times 3 = 18$$
$$6^x = 6$$
$$x = 1$$
x = 1
4. Exercise 9.1 – Q3: Solve
Q3.a: Solve: $4^{\frac{1-x}{1+x}} = 4^{\frac{1}{3}}$
$$\frac{1-x}{1+x} = \frac{1}{3}$$
$$3(1-x) = 1(1+x)$$
$$3 – 3x = 1 + x$$
$$2 = 4x$$
$$x = 1/2$$
$$3(1-x) = 1(1+x)$$
$$3 – 3x = 1 + x$$
$$2 = 4x$$
$$x = 1/2$$
x = 1/2
Q3.b: Solve: $\sqrt[2x+4]{4^{x+8}} = \sqrt[6]{128}$
$4=2^2, 128=2^7$
$$(2^{2(x+8)})^{\frac{1}{2(x+2)}} = (2^7)^{\frac{1}{6}}$$
$$2^{\frac{x+8}{x+2}} = 2^{\frac{7}{6}}$$
$$6(x+8) = 7(x+2)$$
$$6x + 48 = 7x + 14$$
$$34 = x$$
$$2^{\frac{x+8}{x+2}} = 2^{\frac{7}{6}}$$
$$6(x+8) = 7(x+2)$$
$$6x + 48 = 7x + 14$$
$$34 = x$$
x = 34
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Q3.c: Solve: $2^{x+1} + 2^{x+2} + 2^x = 448$
$$2^x(2^1 + 2^2 + 1) = 448$$
$$2^x(7) = 448$$
$$2^x = 64$$
$$2^x = 2^6$$
$$x = 6$$
$$2^x(7) = 448$$
$$2^x = 64$$
$$2^x = 2^6$$
$$x = 6$$
x = 6
Q3.d: Solve: $3^{x+1} – 3^x = 162$
$$3^x(3 – 1) = 162$$
$$3^x(2) = 162$$
$$3^x = 81$$
$$3^x = 3^4$$
$$x = 4$$
$$3^x(2) = 162$$
$$3^x = 81$$
$$3^x = 3^4$$
$$x = 4$$
x = 4
Q3.e: Solve: $4^{x+1} – 8 \times 4^{x-1} = 32$
$$4^{x-1}(4^2 – 8) = 32$$
$$4^{x-1}(8) = 32$$
$$4^{x-1} = 4$$
$$x-1 = 1$$
$$x = 2$$
$$4^{x-1}(8) = 32$$
$$4^{x-1} = 4$$
$$x-1 = 1$$
$$x = 2$$
x = 2
Q3.f: Solve: $4 \times 3^{x+1} – 3^{x+2} – 3^{x-1} = 72$
$$3^{x-1}(4 \cdot 3^2 – 3^3 – 1) = 72$$
$$3^{x-1}(36 – 27 – 1) = 72$$
$$3^{x-1}(8) = 72$$
$$3^{x-1} = 9 = 3^2$$
$$x = 3$$
$$3^{x-1}(36 – 27 – 1) = 72$$
$$3^{x-1}(8) = 72$$
$$3^{x-1} = 9 = 3^2$$
$$x = 3$$
x = 3
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Q3.g: Solve: $3^{x+2} + 3^{x+1} + 2 \times 3^x = 126$
$$3^x(9 + 3 + 2) = 126$$
$$3^x(14) = 126$$
$$3^x = 9$$
$$x = 2$$
$$3^x(14) = 126$$
$$3^x = 9$$
$$x = 2$$
x = 2
Q3.h: Solve: $2^x + 3^{x-2} = 3^x – 2^{x+1}$
$$2^x + 2^{x+1} = 3^x – 3^{x-2}$$
$$2^x(1 + 2) = 3^{x-2}(3^2 – 1)$$
$$2^x(3) = 3^{x-2}(8)$$
$$\frac{2^x}{8} = \frac{3^{x-2}}{3}$$
$$2^{x-3} = 3^{x-3}$$
$$x – 3 = 0$$
$$2^x(1 + 2) = 3^{x-2}(3^2 – 1)$$
$$2^x(3) = 3^{x-2}(8)$$
$$\frac{2^x}{8} = \frac{3^{x-2}}{3}$$
$$2^{x-3} = 3^{x-3}$$
$$x – 3 = 0$$
x = 3
Q3.i: Solve: $8^{x-1} – 2^{3x-2} + 8 = 0$
$$2^{3x-3} – 2^{3x-2} = -8$$
$$2^{3x-3}(1 – 2) = -8$$
$$2^{3x-3}(-1) = -8$$
$$2^{3x-3} = 2^3$$
$$3x = 6$$
$$2^{3x-3}(1 – 2) = -8$$
$$2^{3x-3}(-1) = -8$$
$$2^{3x-3} = 2^3$$
$$3x = 6$$
x = 2
Q3.j: Solve: $(\frac{1}{4})^{2-\sqrt{5x+1}} = 4 \times 2^{\sqrt{5x+1}}$
$$2^{-2(2-\sqrt{5x+1})} = 2^2 \cdot 2^{\sqrt{5x+1}}$$
$$-4 + 2\sqrt{5x+1} = 2 + \sqrt{5x+1}$$
$$\sqrt{5x+1} = 6$$
$$5x + 1 = 36$$
$$x = 7$$
$$-4 + 2\sqrt{5x+1} = 2 + \sqrt{5x+1}$$
$$\sqrt{5x+1} = 6$$
$$5x + 1 = 36$$
$$x = 7$$
x = 7
5. Exercise 9.1 – Q4 to Q7 (Advanced Equations)
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Q4.a: Solve: $5^x + \frac{1}{5^x} = 5\frac{1}{5}$
Let $5^x = a$. $a + 1/a = 26/5$.
$$5a^2 – 26a + 5 = 0$$
$$(5a-1)(a-5) = 0$$
$$a = 5 \implies x = 1$$
$$a = 1/5 \implies x = -1$$
$$(5a-1)(a-5) = 0$$
$$a = 5 \implies x = 1$$
$$a = 1/5 \implies x = -1$$
x = ±1
Q4.b: Solve: $7^x + \frac{1}{7^x} = 7\frac{1}{7}$
Let $7^x = a$. $a + 1/a = 50/7$.
$$a = 7 \implies x = 1$$
$$a = 1/7 \implies x = -1$$
$$a = 1/7 \implies x = -1$$
x = ±1
Q4.c: Solve: $9^x + \frac{1}{9^x} = 9\frac{1}{9}$
Similarly, $x = 1$ or $x = -1$.
x = ±1
Q4.d: Solve: $4^x + \frac{1}{4^x} = 16\frac{1}{16}$
Let $4^x = a$. $a + 1/a = 16 + 1/16$.
$$a = 16 \implies 4^x = 4^2 \implies x = 2$$
$$a = 1/16 \implies 4^x = 4^{-2} \implies x = -2$$
$$a = 1/16 \implies 4^x = 4^{-2} \implies x = -2$$
x = ±2
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Q4.e: Solve: $5^x + 5^{-x} = 25\frac{1}{25}$
Same as Q4.d with base 5.
$$a = 25 \implies x = 2$$
$$a = 1/25 \implies x = -2$$
$$a = 1/25 \implies x = -2$$
x = ±2
Q4.f: Solve: $81 \cdot 3^x + 3^{-x} = 30$
Let $3^x = a$.
$$81a + 1/a = 30$$
$$81a^2 – 30a + 1 = 0$$
$$(9a – 1)^2 = 0$$
$$a = 1/9 \implies 3^x = 3^{-2}$$
$$81a^2 – 30a + 1 = 0$$
$$(9a – 1)^2 = 0$$
$$a = 1/9 \implies 3^x = 3^{-2}$$
x = -2
Q5.a: Solve: $4 \cdot 3^{x+1} – 9^x = 27$
$$12 \cdot 3^x – (3^x)^2 = 27$$
$$Let\ y = 3^x \implies y^2 – 12y + 27 = 0$$
$$(y-3)(y-9) = 0$$
$$y=3 \implies x=1$$
$$y=9 \implies x=2$$
$$Let\ y = 3^x \implies y^2 – 12y + 27 = 0$$
$$(y-3)(y-9) = 0$$
$$y=3 \implies x=1$$
$$y=9 \implies x=2$$
x = 1, 2
Q5.b: Solve: $3 \cdot 2^{p+1} – 4^p = 8$
$$6 \cdot 2^p – (2^p)^2 = 8$$
$$Let\ y = 2^p \implies y^2 – 6y + 8 = 0$$
$$y=2 \implies p=1$$
$$y=4 \implies p=2$$
$$Let\ y = 2^p \implies y^2 – 6y + 8 = 0$$
$$y=2 \implies p=1$$
$$y=4 \implies p=2$$
p = 1, 2
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Q5.c: Solve: $5^{2x} – 6 \cdot 5^{x+1} + 125 = 0$
$$(5^x)^2 – 30 \cdot 5^x + 125 = 0$$
$$y^2 – 30y + 125 = 0$$
$$y=5 \implies x=1$$
$$y=25 \implies x=2$$
$$y^2 – 30y + 125 = 0$$
$$y=5 \implies x=1$$
$$y=25 \implies x=2$$
x = 1, 2
Q5.d: Solve: $2^{x-2} + 2^{3-x} = 3$
$$\frac{y}{4} + \frac{8}{y} = 3 \quad (y=2^x)$$
$$y^2 – 12y + 32 = 0$$
$$y=4 \implies x=2$$
$$y=8 \implies x=3$$
$$y^2 – 12y + 32 = 0$$
$$y=4 \implies x=2$$
$$y=8 \implies x=3$$
x = 2, 3
Q5.e: Solve: $5^{x+1} + 5^{2-x} = 126$
$$5y + \frac{25}{y} = 126 \quad (y=5^x)$$
$$5y^2 – 126y + 25 = 0$$
$$y=25 \implies x=2$$
$$y=1/5 \implies x=-1$$
$$5y^2 – 126y + 25 = 0$$
$$y=25 \implies x=2$$
$$y=1/5 \implies x=-1$$
x = 2, -1
Q5.f: Solve: $3^{2y} – 4 \cdot 3^y + 3 = 0$
$$a^2 – 4a + 3 = 0 \quad (a=3^y)$$
$$a=1 \implies y=0$$
$$a=3 \implies y=1$$
$$a=1 \implies y=0$$
$$a=3 \implies y=1$$
y = 0, 1
Q6: Solve: $16^x – 5 \cdot 4^{x+1} + 64 = 0$
$$(4^x)^2 – 20 \cdot 4^x + 64 = 0$$
$$a^2 – 20a + 64 = 0$$
$$a=4 \implies x=1$$
$$a=16 \implies x=2$$
$$a^2 – 20a + 64 = 0$$
$$a=4 \implies x=1$$
$$a=16 \implies x=2$$
x = 1, 2
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Q7: Solve: $5^x + \frac{125}{5^x} = 30$
$$y + 125/y = 30$$
$$y^2 – 30y + 125 = 0$$
$$y=5 \implies x=1$$
$$y=25 \implies x=2$$
$$y^2 – 30y + 125 = 0$$
$$y=5 \implies x=1$$
$$y=25 \implies x=2$$
x = 1, 2
Q7.a: If $x = 3^{1/3} + 3^{-1/3}$, prove $3x(x^2-3) = 10$
Cube both sides:
$$x^3 = 3 + 1/3 + 3(3^{1/3} \cdot 3^{-1/3})(x)$$
$$x^3 = 10/3 + 3x$$
$$3x^3 = 10 + 9x$$
$$3x^3 – 9x = 10$$
$$3x(x^2 – 3) = 10$$
$$x^3 = 10/3 + 3x$$
$$3x^3 = 10 + 9x$$
$$3x^3 – 9x = 10$$
$$3x(x^2 – 3) = 10$$
Proved
Q7.b: If $x = 2^{1/3} – 2^{-1/3}$, prove $2x^3 + 6x – 3 = 0$
Cube both sides:
$$x^3 = 2 – 1/2 – 3(1)(x)$$
$$x^3 = 3/2 – 3x$$
$$2x^3 = 3 – 6x$$
$$2x^3 + 6x – 3 = 0$$
$$x^3 = 3/2 – 3x$$
$$2x^3 = 3 – 6x$$
$$2x^3 + 6x – 3 = 0$$
Proved
Disclaimer: The solutions provided here are prepared by the Important Edu Notes Team and are based on the CDC curriculum.