Class 9 Adv Maths Ex 4.3

Class 9 / 10 Mathematics: Exponents & Indices

Complete Step-by-Step Solutions (Questions 1 to 19)

Question 1: Express in radical form (মূল চিনত প্রকাশ কৰা)

Express each of the following exponential expressions in radical form:

1. $11^{\frac{2}{3}}$
2. $a^{\frac{1}{4}} b^{\frac{3}{4}}$
3. $x^{\frac{2}{n}} y^{\frac{3}{n}} z^{\frac{4}{n}}$
4. $\left(2^{\frac{3}{4}}\right)^2$
5. $\left(5^6\right)^{\frac{2}{18}}$
6. $\left(m^{\frac{4}{3}} n^{\frac{7}{3}}\right)^{-\frac{1}{5}}$

Solutions

1. $11^{\frac{2}{3}} = (11^2)^{\frac{1}{3}} = \sqrt[3]{11^2} = \mathbf{\sqrt[3]{121}}$
2. $a^{\frac{1}{4}} b^{\frac{3}{4}} = (a \cdot b^3)^{\frac{1}{4}} = \mathbf{\sqrt[4]{a b^3}}$
3. $x^{\frac{2}{n}} y^{\frac{3}{n}} z^{\frac{4}{n}} = (x^2 y^3 z^4)^{\frac{1}{n}} = \mathbf{\sqrt[n]{x^2 y^3 z^4}}$
4. $\left(2^{\frac{3}{4}}\right)^2 = 2^{\frac{3}{4} \times 2} = 2^{\frac{3}{2}} = (2^3)^{\frac{1}{2}} = \sqrt{2^3} = \mathbf{\sqrt{8}}$
5. $\left(5^6\right)^{\frac{2}{18}} = 5^{6 \times \frac{1}{9}} = 5^{\frac{2}{3}} = (5^2)^{\frac{1}{3}} = \mathbf{\sqrt[3]{25}}$
6. $\left(m^{\frac{4}{3}} n^{\frac{7}{3}}\right)^{-\frac{1}{5}} = \left(m^4 n^7\right)^{\frac{1}{3} \times \left(-\frac{1}{5}\right)} = \left(m^4 n^7\right)^{-\frac{1}{15}} = \mathbf{\frac{1}{\sqrt[15]{m^4 n^7}}}$

Question 2: Fill in the blanks (খালী ঠাই পূৰোৱা)

Fill in the blanks to make the equations true:

1. $y^{-\frac{3}{5}} \times \underline{\hspace{2cm}} = 1$
2. $y^7 \times x^{-3} \times \underline{\hspace{2cm}} = y^6 x^2$
3. $x^{\frac{3}{5}} \times y^{\frac{1}{3}} \times \underline{\hspace{2cm}} = y$

Solutions

1. Let the blank be $A$. $y^{-\frac{3}{5}} \times A = 1 \implies A = \frac{1}{y^{-\frac{3}{5}}} = \mathbf{y^{\frac{3}{5}}}$
2. Let the blank be $A$. $y^7 x^{-3} A = y^6 x^2 \implies A = \frac{y^6 x^2}{y^7 x^{-3}} = y^{6-7} x^{2-(-3)} = y^{-1} x^5 = \mathbf{x^5 y^{-1}}$
3. Let the blank be $A$. $x^{\frac{3}{5}} y^{\frac{1}{3}} A = y \implies A = \frac{y}{x^{\frac{3}{5}} y^{\frac{1}{3}}} = x^{-\frac{3}{5}} y^{1 - \frac{1}{3}} = \mathbf{x^{-\frac{3}{5}} y^{\frac{2}{3}}}$

Question 3: Express as a power of 2 (2 ৰ ঘাত হিচাপে লিখা)

Express each of the following as a power of $2$:

1. $\sqrt[3]{64}$
2. $32 \times 8^{n+3}$

Solutions

1. $\sqrt[3]{64} = (64)^{\frac{1}{3}} = (2^6)^{\frac{1}{3}} = 2^{6 \times \frac{1}{3}} = \mathbf{2^2}$
2. $32 \times 8^{n+3} = 2^5 \times (2^3)^{n+3} = 2^5 \times 2^{3(n+3)} = 2^5 \times 2^{3n+9} = 2^{5 + 3n + 9} = \mathbf{2^{3n+14}}$

Question 4: Correct the RHS if wrong (সৌপক্ষ যদি ভুল শুধৰাই লিখা)

Check whether the RHS is correct. If wrong, correct it:

1. $4x^0 = (4x)^0$
2. $1^{-1} = (-1)^1$
3. $-3^1 = (-3)^1$
4. $\left(-\frac{1}{5}\right)^{-5} = \left(\frac{1}{5}\right)^5$
5. $x^{m^n} = x^{mn}$
6. $7^{m-n} = 7^m - 7^n$
7. $(x^{-1} + y^{-1})^{-1} = x^1 + y^1$
8. $\left(5^{-1} + \frac{1}{11}\right)^{-1} = 5 + 11^{-1}$

Solutions

1. Incorrect. $\text{LHS} = 4x^0 = 4(1) = 4$, whereas $\text{RHS} = (4x)^0 = 1$.   Correct RHS: $\mathbf{4}$
2. Incorrect. $\text{LHS} = 1^{-1} = 1$, whereas $\text{RHS} = (-1)^1 = -1$.   Correct RHS: $\mathbf{1}$
3. Correct. $\text{LHS} = -3^1 = -3$ and $\text{RHS} = (-3)^1 = -3$.
4. Incorrect. $\text{LHS} = \left(-\frac{1}{5}\right)^{-5} = (-5)^5 = -3125$, whereas $\text{RHS} = \left(\frac{1}{5}\right)^5 = \frac{1}{3125}$.   Correct RHS: $\mathbf{-3125}$
5. Incorrect. $x^{m^n}$ means $x^{(m^n)}$. The identity $(x^m)^n = x^{mn}$ applies only when $x^m$ is in brackets.   Correct Identity: $\mathbf{(x^m)^n = x^{mn}}$
6. Incorrect. By quotient rule of exponents, $7^{m-n} = \frac{7^m}{7^n} \neq 7^m - 7^n$.   Correct RHS: $\mathbf{\frac{7^m}{7^n}}$
7. Incorrect. $\text{LHS} = \left(\frac{1}{x} + \frac{1}{y}\right)^{-1} = \left(\frac{x+y}{xy}\right)^{-1} = \frac{xy}{x+y}$.   Correct RHS: $\mathbf{\frac{xy}{x+y}}$
8. Incorrect. $\text{LHS} = \left(\frac{1}{5} + \frac{1}{11}\right)^{-1} = \left(\frac{11+5}{55}\right)^{-1} = \left(\frac{16}{55}\right)^{-1} = \frac{55}{16}$.   Correct RHS: $\mathbf{\frac{55}{16}}$

Question 5: Evaluate / Find the Value (মান নিৰ্ণয় কৰা)

Evaluate each of the following expressions:

1. $\sqrt[5]{a^{15} b^{10} c^5}$
2. $(\sqrt{x})^{\frac{1}{2}} \sqrt{y^4} \div \sqrt{x y^{-\frac{1}{2}}}$
3. $\left(\frac{2y^3 a^2}{3a^3 y^2}\right) \div \left(\frac{3a}{2y}\right)^3$
4. $\left(\frac{x^{-1} y^2}{x^2 y^{-4}}\right)^3 \div \left(\frac{x^3 y^{-5}}{x^{-2} y^3}\right)^{-5}$
5. $\left(\sqrt{3 \times 5^{-3}} \div \sqrt[3]{3^{-1} \sqrt{5}}\right) \times \sqrt[6]{3 \times 5^6}$
6. $\frac{3^m + 3^{m-1}}{3^{m+1} - 3^m}$
7. $\frac{m^{-2} n^2}{m^{-1} + n^{-1}} \div \frac{1}{m(m+n)}$
8. $\frac{16 \times 2^{m+1} - 4 \times 2^m}{16 \times 2^{m+2} - 2 \times 2^{m+3}}$

Solutions

1. $\sqrt[5]{a^{15} b^{10} c^5} = \left(a^{15} b^{10} c^5\right)^{\frac{1}{5}} = a^{\frac{15}{5}} b^{\frac{10}{5}} c^{\frac{5}{5}} = \mathbf{a^3 b^2 c}$
2. $\frac{(x^{1/2})^{1/2} \cdot (y^4)^{1/2}}{(x \cdot y^{-1/2})^{1/2}} = \frac{x^{1/4} y^2}{x^{1/2} y^{-1/4}} = x^{\frac{1}{4}-\frac{1}{2}} y^{2 - (-\frac{1}{4})} = x^{-\frac{1}{4}} y^{\frac{9}{4}} = \mathbf{\frac{y^{9/4}}{x^{1/4}}}$
3. $\left(\frac{2y}{3a}\right) \div \left(\frac{27a^3}{8y^3}\right) = \frac{2y}{3a} \times \frac{8y^3}{27a^3} = \mathbf{\frac{16y^4}{81a^4}}$
4. $\left(x^{-3} y^6\right)^3 \div \left(x^5 y^{-8}\right)^{-5} = \left(x^{-9} y^{18}\right) \div \left(x^{-25} y^{40}\right) = x^{-9 - (-25)} y^{18 - 40} = x^{16} y^{-22} = \mathbf{\frac{x^{16}}{y^{22}}}$
5. Express all terms as powers of $3$ and $5$: $$\text{Term 1} = 3^{\frac{1}{2}} 5^{-\frac{3}{2}}, \quad \text{Term 2} = \left(3^{-1} 5^{\frac{1}{2}}\right)^{\frac{1}{3}} = 3^{-\frac{1}{3}} 5^{\frac{1}{6}}, \quad \text{Term 3} = 3^{\frac{1}{6}} 5^{1}$$ $$\text{Expression} = \frac{3^{\frac{1}{2}} 5^{-\frac{3}{2}}}{3^{-\frac{1}{3}} 5^{\frac{1}{6}}} \times 3^{\frac{1}{6}} 5^{1} = 3^{\frac{1}{2} + \frac{1}{3} + \frac{1}{6}} \cdot 5^{-\frac{3}{2} - \frac{1}{6} + 1} = 3^1 \cdot 5^{-\frac{2}{3}} = \mathbf{\frac{3}{\sqrt[3]{25}}}$$
6. Factor out $3^m$: $$\frac{3^m \left(1 + 3^{-1}\right)}{3^m \left(3^1 - 1\right)} = \frac{1 + \frac{1}{3}}{2} = \frac{\frac{4}{3}}{2} = \mathbf{\frac{2}{3}}$$
7. Simplify numerator and denominator: $$\frac{\frac{n^2}{m^2}}{\frac{1}{m} + \frac{1}{n}} \times m(m+n) = \frac{\frac{n^2}{m^2}}{\frac{m+n}{mn}} \times m(m+n) = \frac{n^2}{m^2} \times \frac{mn}{m+n} \times m(m+n) = \mathbf{n^3}$$
8. Express numbers as powers of 2: $$\frac{2^4 \cdot 2^{m+1} - 2^2 \cdot 2^m}{2^4 \cdot 2^{m+2} - 2^1 \cdot 2^{m+3}} = \frac{2^{m+5} - 2^{m+2}}{2^{m+6} - 2^{m+4}} = \frac{2^{m+2}\left(2^3 - 1\right)}{2^{m+4}\left(2^2 - 1\right)} = \frac{2^{m+2} \times 7}{2^{m+4} \times 3} = \frac{7}{2^2 \times 3} = \mathbf{\frac{7}{12}}$$

Question 6: Simplify (সৰল কৰা)

Simplify the following expressions:

1. $\sqrt[5]{\sqrt[5]{\sqrt{x^{625}}}}$
2. $\left\{x^2 \sqrt{49 \sqrt[3]{x^6 y^{-12}}}\right\}^{\frac{1}{2}}$
3. $9^{\frac{3}{2}} - 3(5)^0 - \left(\frac{1}{81}\right)^{-\frac{1}{2}}$
4. $\frac{27^{\frac{2n}{3}} \times 8^{-\frac{n}{6}}}{(18)^{-\frac{n}{2}}}$
5. $\left(\frac{a^m}{a^n}\right)^{m+n} \times \left(\frac{a^n}{a^l}\right)^{n+l} \times \left(\frac{a^l}{a^m}\right)^{l+m}$
6. $\left(\frac{a^l}{a^m}\right)^{l^2 + lm + m^2}$
7. $\left(\frac{a^m}{a^n}\right)^l \times \left(\frac{a^n}{a^l}\right)^m \times \left(\frac{a^l}{a^m}\right)^n$
8. $\frac{6^n \times 6^{2n} \times 5^{3n}}{30^n \times 3^{2n} \times 2^{2n}}$

Solutions

1. $\left(x^{625}\right)^{\frac{1}{2} \times \frac{1}{5} \times \frac{1}{5}} = \left(x^{625}\right)^{\frac{1}{50}} = x^{\frac{625}{50}} = \mathbf{x^{\frac{25}{2}}}$
2. Inner cube root: $\sqrt[3]{x^6 y^{-12}} = x^2 y^{-4}$.
   Square root term: $\sqrt{49 x^2 y^{-4}} = 7 x y^{-2}$.
   Whole expression: $\left\{x^2 \cdot 7x y^{-2}\right\}^{\frac{1}{2}} = \left\{7 x^3 y^{-2}\right\}^{\frac{1}{2}} = \mathbf{\frac{\sqrt{7} x^{3/2}}{y}}$
3. $(3^2)^{\frac{3}{2}} - 3(1) - (81)^{\frac{1}{2}} = 3^3 - 3 - 9 = 27 - 12 = \mathbf{15}$
4. Convert to prime bases: $$\frac{(3^3)^{\frac{2n}{3}} \times (2^3)^{-\frac{n}{6}}}{(2 \cdot 3^2)^{-\frac{n}{2}}} = \frac{3^{2n} \times 2^{-\frac{n}{2}}}{2^{-\frac{n}{2}} \times 3^{-n}} = 3^{2n - (-n)} = \mathbf{3^{3n} \quad (\text{or } 27^n)}$$
5. $(a^{m-n})^{m+n} \cdot (a^{n-l})^{n+l} \cdot (a^{l-m})^{l+m} = a^{m^2-n^2} \cdot a^{n^2-l^2} \cdot a^{l^2-m^2} = a^{m^2-n^2+n^2-l^2+l^2-m^2} = a^0 = \mathbf{1}$
6. $(a^{l-m})^{l^2 + lm + m^2} = a^{(l-m)(l^2+lm+m^2)} = \mathbf{a^{l^3 - m^3}}$
7. $a^{(m-n)l} \cdot a^{(n-l)m} \cdot a^{(l-m)n} = a^{lm - nl + mn - lm + nl - mn} = a^0 = \mathbf{1}$
8. Express numerator and denominator in terms of powers of 2, 3, 5: $$\text{Numerator} = 6^{3n} \times 5^{3n} = 2^{3n} \cdot 3^{3n} \cdot 5^{3n}$$ $$\text{Denominator} = (2 \cdot 3 \cdot 5)^n \times 3^{2n} \times 2^{2n} = 2^{3n} \cdot 3^{3n} \cdot 5^n$$ $$\text{Result} = \frac{2^{3n} \cdot 3^{3n} \cdot 5^{3n}}{2^{3n} \cdot 3^{3n} \cdot 5^n} = 5^{3n-n} = \mathbf{5^{2n} \quad (\text{or } 25^n)}$$

Question 7: Prove that (প্রমাণ কৰা)

Prove the following identities:

1. $\left(\frac{x^a}{x^b}\right)^{a^2+ab+b^2} \times \left(\frac{x^b}{x^c}\right)^{b^2+bc+c^2} \times \left(\frac{x^c}{x^a}\right)^{c^2+ca+a^2} = 1$
2. $\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} \times \left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}} \times \left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}} = 1$
3. $\sqrt[b+c]{\frac{x^{b^2}}{x^{c^2}}} \times \sqrt[c+a]{\frac{x^{c^2}}{x^{a^2}}} \times \sqrt[a+b]{\frac{x^{a^2}}{x^{b^2}}} = 1$
4. $\frac{1}{a^l + a^{-m} + 1} + \frac{1}{a^m + a^{-n} + 1} + \frac{1}{a^n + a^{-l} + 1} = 1 \quad \text{when } l+m+n = 0$
5. $\frac{\left(x + \frac{1}{y}\right)^m \left(x - \frac{1}{y}\right)^m}{\left(y + \frac{1}{x}\right)^m \left(y - \frac{1}{x}\right)^m} = \left(\frac{x}{y}\right)^{2m}$
6. $\left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}} \times \left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}} \times \left(\frac{x^c}{x^a}\right)^{\frac{1}{ca}} = 1$
7. $\frac{x+y+z}{x^{-1}y^{-1} + y^{-1}z^{-1} + z^{-1}x^{-1}} = xyz$
8. $\frac{(x^{l+m})^2 \times (x^{m+n})^2 \times (x^{n+l})^2}{(x^l x^m x^n)^4} = 1$

Solutions for Question 7

1. $\text{LHS} = x^{(a-b)(a^2+ab+b^2)} \cdot x^{(b-c)(b^2+bc+c^2)} \cdot x^{(c-a)(c^2+ca+a^2)} = x^{a^3-b^3} \cdot x^{b^3-c^3} \cdot x^{c^3-a^3} = x^{a^3-b^3+b^3-c^3+c^3-a^3} = x^0 = 1 = \text{RHS}$. ■
2. Exponent sum: $$S = \frac{1}{(a-b)(a-c)} + \frac{1}{(b-c)(b-a)} + \frac{1}{(c-a)(c-b)} = \frac{-1}{(a-b)(c-a)} - \frac{1}{(a-b)(b-c)} - \frac{1}{(b-c)(c-a)}$$ Taking LCM $(a-b)(b-c)(c-a)$: $$S = \frac{-(b-c) - (c-a) - (a-b)}{(a-b)(b-c)(c-a)} = \frac{-b+c-c+a-a+b}{(a-b)(b-c)(c-a)} = \frac{0}{(a-b)(b-c)(c-a)} = 0$$ $\implies \text{LHS} = x^0 = 1 = \text{RHS}$. ■
3. $\text{LHS} = \left(x^{b^2-c^2}\right)^{\frac{1}{b+c}} \cdot \left(x^{c^2-a^2}\right)^{\frac{1}{c+a}} \cdot \left(x^{a^2-b^2}\right)^{\frac{1}{a+b}} = x^{\frac{(b-c)(b+c)}{b+c}} \cdot x^{\frac{(c-a)(c+a)}{c+a}} \cdot x^{\frac{(a-b)(a+b)}{a+b}} = x^{b-c} \cdot x^{c-a} \cdot x^{a-b} = x^0 = 1 = \text{RHS}$. ■
4. Given $l+m+n = 0 \implies n = -(l+m)$.
   $\text{Term 1} = \frac{1}{a^l + a^{-m} + 1}$
   $\text{Term 2} = \frac{1}{a^m + a^{l+m} + 1} = \frac{a^{-m}}{1 + a^l + a^{-m}}$
   $\text{Term 3} = \frac{1}{a^{-(l+m)} + a^{-l} + 1} = \frac{a^l}{a^{-m} + 1 + a^l}$
   $\text{LHS} = \frac{1 + a^{-m} + a^l}{a^l + a^{-m} + 1} = 1 = \text{RHS}$. ■
5. $\text{LHS} = \frac{\left[\left(x+\frac{1}{y}\right)\left(x-\frac{1}{y}\right)\right]^m}{\left[\left(y+\frac{1}{x}\right)\left(y-\frac{1}{x}\right)\right]^m} = \frac{\left(x^2 - \frac{1}{y^2}\right)^m}{\left(y^2 - \frac{1}{x^2}\right)^m} = \frac{\left(\frac{x^2 y^2 - 1}{y^2}\right)^m}{\left(\frac{x^2 y^2 - 1}{x^2}\right)^m} = \frac{\frac{(x^2 y^2 - 1)^m}{y^{2m}}}{\frac{(x^2 y^2 - 1)^m}{x^{2m}}} = \frac{x^{2m}}{y^{2m}} = \left(\frac{x}{y}\right)^{2m} = \text{RHS}$. ■
6. $\text{LHS} = x^{\frac{a-b}{ab}} \cdot x^{\frac{b-c}{bc}} \cdot x^{\frac{c-a}{ca}} = x^{\left(\frac{1}{b} - \frac{1}{a}\right) + \left(\frac{1}{c} - \frac{1}{b}\right) + \left(\frac{1}{a} - \frac{1}{c}\right)} = x^0 = 1 = \text{RHS}$. ■
7. $\text{LHS} = \frac{x+y+z}{\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}} = \frac{x+y+z}{\frac{z+x+y}{xyz}} = (x+y+z) \times \frac{xyz}{x+y+z} = xyz = \text{RHS}$. ■
8. $\text{LHS} = \frac{x^{2(l+m)} \cdot x^{2(m+n)} \cdot x^{2(n+l)}}{\left(x^{l+m+n}\right)^4} = \frac{x^{2l+2m+2m+2n+2n+2l}}{x^{4l+4m+4n}} = \frac{x^{4l+4m+4n}}{x^{4l+4m+4n}} = 1 = \text{RHS}$. ■

Question 8

If $(x^m)^n = x^{m^n}$, show that $m^{n-1} \times n^{m-1} = n^m$.

Proof

Given $(x^m)^n = x^{m^n} \implies x^{mn} = x^{m^n}$.

Equating exponents: $$mn = m^n \implies n = \frac{m^n}{m} = m^{n-1}$$

Now evaluate LHS: $$\text{LHS} = m^{n-1} \times n^{m-1} = n \times n^{m-1} = n^{1 + (m-1)} = n^m = \text{RHS} \quad \span class="proof-end">■$$

Question 9

If $\frac{9^n \times 3^2 \times (3^{-n})^{-1} - (27)^n}{3^{3m} \times 2^3} = \frac{1}{27}$, show that $m = 1+n$.

Proof

$$\text{LHS} = \frac{(3^2)^n \times 3^2 \times 3^n - (3^3)^n}{3^{3m} \times 8} = \frac{3^{2n+2+n} - 3^{3n}}{3^{3m} \times 8} = \frac{3^{3n+2} - 3^{3n}}{3^{3m} \times 8}$$

Factor out $3^{3n}$:

$$\text{LHS} = \frac{3^{3n}(3^2 - 1)}{3^{3m} \times 8} = \frac{3^{3n} \times 8}{3^{3m} \times 8} = 3^{3n - 3m}$$

Given LHS $= \frac{1}{27} = 3^{-3}$:

$$3^{3n - 3m} = 3^{-3} \implies 3n - 3m = -3 \implies 3m - 3n = 3 \implies m - n = 1 \implies \mathbf{m = 1 + n} \quad \span class="proof-end">■$$

Question 10

If $a+b+c = 0$, show that $\frac{1}{x^b + x^{-c} + 1} + \frac{1}{x^c + x^{-a} + 1} + \frac{1}{x^a + x^{-b} + 1} = 1$.

Proof

Given $a+b+c = 0 \implies -c = a+b$, $-a = b+c$, $-b = a+c$.

Rewrite each term with common denominator $(x^a + x^{-b} + 1)$:

$$\text{Term 1} = \frac{1}{x^b + x^{a+b} + 1} = \frac{x^{-b}}{1 + x^a + x^{-b}}$$ $$\text{Term 2} = \frac{1}{x^c + x^{b+c} + 1} = \frac{x^a}{x^{a+c} + x^{a+b+c} + x^a} = \frac{x^a}{x^{-b} + 1 + x^a} = \frac{x^a}{x^a + x^{-b} + 1}$$ $$\text{Term 3} = \frac{1}{x^a + x^{-b} + 1}$$

Summing all terms:

$$\text{LHS} = \frac{x^{-b} + x^a + 1}{x^a + x^{-b} + 1} = 1 = \text{RHS} \quad \span class="proof-end">■$$

Question 11

If $a = b^x$, $b = c^y$, $c = a^z$, show that $xyz = 1$.

Proof

Given $a = b^x$. Substitute $b = c^y$:

$$a = (c^y)^x = c^{xy}$$

Now substitute $c = a^z$:

$$a = (a^z)^{xy} = a^{xyz}$$

Since $a^1 = a^{xyz}$, equating exponents gives $\mathbf{xyz = 1}$. ■

Question 12

If $a^x = b^y = c^z$ and $b^2 = ac$, show that $\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$, and hence show $y = \frac{2zx}{z+x}$.

Proof

Let $a^x = b^y = c^z = k$. Then $a = k^{1/x}$, $b = k^{1/y}$, $c = k^{1/z}$.

Substitute into $b^2 = ac$:

$$(k^{1/y})^2 = k^{1/x} \cdot k^{1/z} \implies k^{2/y} = k^{\frac{1}{x} + \frac{1}{z}}$$

Equating exponents:

$$\mathbf{\frac{1}{x} + \frac{1}{z} = \frac{2}{y}} \quad \span class="proof-end">■$$

Solving for $y$:

$$\frac{2}{y} = \frac{z + x}{zx} \implies \mathbf{y = \frac{2zx}{z+x}} \quad \span class="proof-end">■$$

Question 13

If $2^x = 3^y = 6^{-z}$, show that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$.

Proof

Let $2^x = 3^y = 6^{-z} = k \implies 2 = k^{1/x}, \; 3 = k^{1/y}, \; 6 = k^{-1/z}$.

Since $2 \times 3 = 6$:

$$k^{1/x} \times k^{1/y} = k^{-1/z} \implies k^{\frac{1}{x} + \frac{1}{y}} = k^{-\frac{1}{z}}$$

Equating exponents:

$$\frac{1}{x} + \frac{1}{y} = -\frac{1}{z} \implies \mathbf{\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0} \quad \span class="proof-end">■$$

Question 14

If $a = x^{q+r} y^p$, $b = x^{r+p} y^q$, $c = x^{p+q} y^r$, prove that $a^{q-r} \times b^{r-p} \times c^{p-q} = 1$.

Proof

Compute each term individually:

$$a^{q-r} = \left(x^{q+r} y^p\right)^{q-r} = x^{(q+r)(q-r)} y^{p(q-r)} = x^{q^2 - r^2} y^{pq - pr}$$ $$b^{r-p} = \left(x^{r+p} y^q\right)^{r-p} = x^{(r+p)(r-p)} y^{q(r-p)} = x^{r^2 - p^2} y^{qr - qp}$$ $$c^{p-q} = \left(x^{p+q} y^r\right)^{p-q} = x^{(p+q)(p-q)} y^{r(p-q)} = x^{p^2 - q^2} y^{rp - rq}$$

Multiplying all three terms:

$$\text{LHS} = x^{(q^2 - r^2) + (r^2 - p^2) + (p^2 - q^2)} \cdot y^{(pq - pr) + (qr - qp) + (rp - rq)} = x^0 \cdot y^0 = 1 = \text{RHS} \quad \span class="proof-end">■$$

Question 15

If $x = 2 - 3^{\frac{1}{3}}$, show that $x^3 - 6x^2 + 12x - 5 = 0$.

Proof

Rearrange the equation: $2 - x = 3^{\frac{1}{3}}$.

Cubing both sides:

$$(2 - x)^3 = \left(3^{\frac{1}{3}}\right)^3 \implies 2^3 - 3(2)^2 x + 3(2) x^2 - x^3 = 3$$ $$8 - 12x + 6x^2 - x^3 = 3 \implies x^3 - 6x^2 + 12x - 8 + 3 = 0 \implies \mathbf{x^3 - 6x^2 + 12x - 5 = 0} \quad \span class="proof-end">■$$

Question 16

If $x = 3^{\frac{1}{3}} + 3^{-\frac{1}{3}}$, show that $3x^3 - 9x = 10$.

Proof

Cubing both sides using $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$:

$$x^3 = \left(3^{\frac{1}{3}}\right)^3 + \left(3^{-\frac{1}{3}}\right)^3 + 3\left(3^{\frac{1}{3}}\right)\left(3^{-\frac{1}{3}}\right)\left(3^{\frac{1}{3}} + 3^{-\frac{1}{3}}\right)$$ $$x^3 = 3 + 3^{-1} + 3(1)(x) = 3 + \frac{1}{3} + 3x \implies x^3 - 3x = \frac{10}{3} \implies \mathbf{3x^3 - 9x = 10} \quad \span class="proof-end">■$$

Question 17

If $x = \frac{1}{3}(\sqrt{2} - 1)$, show that $9x^3 + 24x^2 + 11x - 2 = 0$.

Proof

Given $3x = \sqrt{2} - 1 \implies 3x + 1 = \sqrt{2}$.

Squaring both sides:

$$(3x + 1)^2 = 2 \implies 9x^2 + 6x + 1 = 2 \implies 9x^2 + 6x - 1 = 0$$

Now factor the target cubic expression:

$$9x^3 + 24x^2 + 11x - 2 = x\left(9x^2 + 6x - 1\right) + 2\left(9x^2 + 6x - 1\right) = (x + 2)\left(9x^2 + 6x - 1\right)$$

Since $9x^2 + 6x - 1 = 0$, we have $(x+2)(0) = \mathbf{0}$. ■

Question 18

If $p = 2^{\frac{2}{3}} + 2^{\frac{1}{3}} + 1$, show that $p^3 - 3p^2 - 3p - 1 = 0$.

Proof

Rearrange: $p - 1 = 2^{\frac{2}{3}} + 2^{\frac{1}{3}}$.

Cubing both sides:

$$(p - 1)^3 = \left(2^{\frac{2}{3}} + 2^{\frac{1}{3}}\right)^3$$ $$p^3 - 3p^2 + 3p - 1 = \left(2^{\frac{2}{3}}\right)^3 + \left(2^{\frac{1}{3}}\right)^3 + 3\left(2^{\frac{2}{3}}\right)\left(2^{\frac{1}{3}}\right)\left(2^{\frac{2}{3}} + 2^{\frac{1}{3}}\right)$$ $$p^3 - 3p^2 + 3p - 1 = 2^2 + 2^1 + 3(2)(p - 1) = 4 + 2 + 6p - 6 = 6p$$ $$p^3 - 3p^2 + 3p - 1 - 6p = 0 \implies \mathbf{p^3 - 3p^2 - 3p - 1 = 0} \quad \span class="proof-end">■$$

Question 19: Solve the following equations (সমাধান কৰা)

Solve each of the following exponential equations for the unknown variable:

1. $9 \times 81^x = \frac{1}{27^{x-3}}$
2. $4^{y-1} = 2^{y-5}$
3. $4^{x+2} + 2^{2x+1} = 36$
4. $3^x = 4^{-x}$
5. $2^{5x} = 64 \times 4^x$
6. $(\sqrt{3})^{\frac{x+1}{2}} = (\sqrt[3]{3})^{2x+3}$
7. $5^{3x+1} = (\sqrt{5})^{x+3}$
8. $x^y = y^x, \quad x = 2y$

Solutions

1. $3^2 \times (3^4)^x = (3^3)^{-(x-3)} \implies 3^{4x+2} = 3^{-3x+9}$
   Equating exponents: $4x + 2 = -3x + 9 \implies 7x = 7 \implies \mathbf{x = 1}$
2. $(2^2)^{y-1} = 2^{y-5} \implies 2^{2y-2} = 2^{y-5}$
   Equating exponents: $2y - 2 = y - 5 \implies \mathbf{y = -3}$
3. $(2^2)^{x+2} + 2^{2x+1} = 36 \implies 2^{2x+4} + 2^{2x+1} = 36$
   $16 \cdot 2^{2x} + 2 \cdot 2^{2x} = 36 \implies 18 \cdot 2^{2x} = 36 \implies 2^{2x} = 2^1 \implies 2x = 1 \implies \mathbf{x = \frac{1}{2}}$
4. $3^x = \frac{1}{4^x} \implies 3^x \cdot 4^x = 1 \implies 12^x = 12^0 \implies \mathbf{x = 0}$
5. $2^{5x} = 2^6 \times (2^2)^x = 2^{6 + 2x}$
   Equating exponents: $5x = 6 + 2x \implies 3x = 6 \implies \mathbf{x = 2}$
6. $\left(3^{\frac{1}{2}}\right)^{\frac{x+1}{2}} = \left(3^{\frac{1}{3}}\right)^{2x+3} \implies 3^{\frac{x+1}{4}} = 3^{\frac{2x+3}{3}}$
   Equating exponents: $\frac{x+1}{4} = \frac{2x+3}{3} \implies 3(x+1) = 4(2x+3) \implies 3x+3 = 8x+12 \implies 5x = -9 \implies \mathbf{x = -\frac{9}{5}}$
7. $5^{3x+1} = \left(5^{\frac{1}{2}}\right)^{x+3} \implies 5^{3x+1} = 5^{\frac{x+3}{2}}$
   Equating exponents: $3x + 1 = \frac{x+3}{2} \implies 6x + 2 = x + 3 \implies 5x = 1 \implies \mathbf{x = \frac{1}{5}}$
8. Substitute $x = 2y$ into $x^y = y^x$: $$(2y)^y = y^{2y}$$ Taking $y$-th root on both sides ($y \neq 0$): $$2y = y^2 \implies y^2 - 2y = 0 \implies y(y-2) = 0 \implies y = 2$$ Since $x = 2y$: $x = 2(2) = 4$.
   Thus, $\mathbf{x = 4, \quad y = 2}$.