Exercise 4.1: Exponents
Simplify (1 – 8):
1. $\frac{7^3 \times 7^{-3}}{3 \times 3^{-4}}$
Solution: $\frac{7^3 \times 7^{-3}}{3 \times 3^{-4}}$
$= \frac{7^{3 + (-3)}}{3^{1 + (-4)}}$
$= \frac{7^0}{3^{-3}}$
$= \frac{1}{\frac{1}{3^3}}$
$= 1 \times 3^3 = 27$
Answer: $27$
2. $\frac{\sqrt[3]{7^2} \cdot \sqrt[3]{7}}{\sqrt{7}}$
Solution: $\frac{\sqrt[3]{7^2} \cdot \sqrt[3]{7}}{\sqrt{7}}$
$= \frac{7^{\frac{2}{3}} \cdot 7^{\frac{1}{3}}}{7^{\frac{1}{2}}}$
$= \frac{7^{\frac{2}{3} + \frac{1}{3}}}{7^{\frac{1}{2}}}$
$= \frac{7^{\frac{3}{3}}}{7^{\frac{1}{2}}} = \frac{7^1}{7^{\frac{1}{2}}}$
$= 7^{1 – \frac{1}{2}} = 7^{\frac{1}{2}} = \sqrt{7}$
Answer: $\sqrt{7}$
3. $(2^{-1} + 5^{-1})^{-1}$
Solution: $(2^{-1} + 5^{-1})^{-1}$
$= \left(\frac{1}{2} + \frac{1}{5}\right)^{-1}$
$= \left(\frac{5 + 2}{10}\right)^{-1}$
$= \left(\frac{7}{10}\right)^{-1} = \frac{10}{7}$
Answer: $\frac{10}{7}$
4. $(2a^{-1} + 3b^{-1})^{-1}$
Solution: $(2a^{-1} + 3b^{-1})^{-1}$
$= \left(\frac{2}{a} + \frac{3}{b}\right)^{-1}$
$= \left(\frac{2b + 3a}{ab}\right)^{-1}$
$= \frac{ab}{3a + 2b}$
Answer: $\frac{ab}{3a + 2b}$
5. $\left(\frac{a^2 b^{-1}}{a^{-2} b}\right)^2$
Solution: $\left(\frac{a^2 b^{-1}}{a^{-2} b}\right)^2$
$= \left(a^{2 – (-2)} \cdot b^{-1 – 1}\right)^2$
$= \left(a^4 \cdot b^{-2}\right)^2$
$= \left(\frac{a^4}{b^2}\right)^2$
$= \frac{a^8}{b^4}$
Answer: $\frac{a^8}{b^4}$
6. $\sqrt{x^{-1}y} \cdot \sqrt{y^{-1}z} \cdot \sqrt{z^{-1}x} \quad (x > 0, y > 0, z > 0)$
Solution: $\sqrt{x^{-1}y} \cdot \sqrt{y^{-1}z} \cdot \sqrt{z^{-1}x} \quad (x > 0, y > 0, z > 0)$
$= \sqrt{\frac{y}{x}} \cdot \sqrt{\frac{z}{y}} \cdot \sqrt{\frac{x}{z}}$
$= \sqrt{\frac{y}{x} \cdot \frac{z}{y} \cdot \frac{x}{z}}$
$= \sqrt{1} = 1$
Answer: $1$
7. $\frac{2^{n+4} – 4 \cdot 2^{n+1}}{2^{n+2} \div 2}$
Solution: $\frac{2^{n+4} – 4 \cdot 2^{n+1}}{2^{n+2} \div 2}$
$= \frac{2^n \cdot 2^4 – 2^2 \cdot 2^{n+1}}{2^{n+2 – 1}}$
$= \frac{2^n \cdot 16 – 2^{n+3}}{2^{n+1}}$
$= \frac{2^n \cdot 16 – 2^n \cdot 2^3}{2^n \cdot 2^1}$
$= \frac{2^n (16 – 8)}{2^n \cdot 2}$
$= \frac{8}{2} = 4$
Answer: $4$
8. $\frac{3^{m+1}}{(3^m)^{m-1}} \div \frac{9^{m+1}}{(3^{m-1})^{m+1}}$
Solution: $\frac{3^{m+1}}{(3^m)^{m-1}} \div \frac{9^{m+1}}{(3^{m-1})^{m+1}}$
$= \frac{3^{m+1}}{3^{m^2 – m}} \div \frac{(3^2)^{m+1}}{3^{(m-1)(m+1)}}$
$= 3^{(m+1) – (m^2 – m)} \div \frac{3^{2m+2}}{3^{m^2 – 1}}$
$= 3^{m+1 – m^2 + m} \div 3^{(2m+2) – (m^2 – 1)}$
$= 3^{2m + 1 – m^2} \div 3^{2m + 2 – m^2 + 1}$
$= 3^{2m + 1 – m^2} \div 3^{2m + 3 – m^2}$
$= 3^{(2m + 1 – m^2) – (2m + 3 – m^2)}$
$= 3^{2m + 1 – m^2 – 2m – 3 + m^2}$
$= 3^{-2} = \frac{1}{3^2} = \frac{1}{9}$
Answer: $\frac{1}{9}$
Prove (9 – 15):
9. $\frac{4^n – 1}{2^n – 1} = 2^n + 1$
Solution: $\text{L.H.S.} = \frac{4^n – 1}{2^n – 1}$
$= \frac{(2^2)^n – 1}{2^n – 1}$
$= \frac{(2^n)^2 – 1^2}{2^n – 1}$
$= \frac{(2^n + 1)(2^n – 1)}{2^n – 1}$
$= 2^n + 1 = \text{R.H.S.}$
$\therefore \frac{4^n – 1}{2^n – 1} = 2^n + 1$ (Proved)
10. $\frac{2^{2p+1} \cdot 3^{2p+q} \cdot 5^{p+q} \cdot 6^p}{3^{p-2} \cdot 6^{2p+2} \cdot 10^p \cdot 15^q} = \frac{1}{2}$
Solution:
$\text{L.H.S.} = \frac{2^{2p+1} \cdot 3^{2p+q} \cdot 5^{p+q} \cdot 6^p}{3^{p-2} \cdot 6^{2p+2} \cdot 10^p \cdot 15^q}$
$= \frac{2^{2p+1} \cdot 3^{2p+q} \cdot 5^{p+q} \cdot (2 \cdot 3)^p}{3^{p-2} \cdot (2 \cdot 3)^{2p+2} \cdot (2 \cdot 5)^p \cdot (3 \cdot 5)^q}$
$= \frac{2^{2p+1} \cdot 3^{2p+q} \cdot 5^{p+q} \cdot 2^p \cdot 3^p}{3^{p-2} \cdot 2^{2p+2} \cdot 3^{2p+2} \cdot 2^p \cdot 5^p \cdot 3^q \cdot 5^q}$
$= \frac{2^{(2p+1)+p} \cdot 3^{(2p+q)+p} \cdot 5^{p+q}}{2^{(2p+2)+p} \cdot 3^{(p-2)+(2p+2)+q} \cdot 5^{p+q}}$
$= \frac{2^{3p+1} \cdot 3^{3p+q} \cdot 5^{p+q}}{2^{3p+2} \cdot 3^{3p+q} \cdot 5^{p+q}}$
$= 2^{(3p+1) – (3p+2)} \cdot 3^{(3p+q) – (3p+q)} \cdot 5^{(p+q) – (p+q)}$
$= 2^{3p + 1 – 3p – 2} \cdot 3^0 \cdot 5^0$
$= 2^{-1} \cdot 1 \cdot 1 \quad [\because a^0 = 1]$
$= \frac{1}{2} = \text{R.H.S.}$
$\therefore \frac{2^{2p+1} \cdot 3^{2p+q} \cdot 5^{p+q} \cdot 6^p}{3^{p-2} \cdot 6^{2p+2} \cdot 10^p \cdot 15^q} = \frac{1}{2}$ (Proved)
11. $\left(\frac{a^l}{a^m}\right)^n \cdot \left(\frac{a^m}{a^n}\right)^l \cdot \left(\frac{a^n}{a^l}\right)^m = 1$
Solution:
$\text{L.H.S.} = \left(\frac{a^l}{a^m}\right)^n \cdot \left(\frac{a^m}{a^n}\right)^l \cdot \left(\frac{a^n}{a^l}\right)^m$
$= (a^{l-m})^n \cdot (a^{m-n})^l \cdot (a^{n-l})^m$
$= a^{ln – mn} \cdot a^{lm – ln} \cdot a^{mn – lm}$
$= a^{ln – mn + lm – ln + mn – lm}$
$= a^0 = 1 = \text{R.H.S.}$
$\therefore \left(\frac{a^l}{a^m}\right)^n \cdot \left(\frac{a^m}{a^n}\right)^l \cdot \left(\frac{a^n}{a^l}\right)^m = 1$ (Proved)
12. $\frac{a^{p+q}}{a^{2r}} \times \frac{a^{q+r}}{a^{2p}} \times \frac{a^{r+p}}{a^{2q}} = 1$
Solution:
$\text{L.H.S.} = \frac{a^{p+q}}{a^{2r}} \times \frac{a^{q+r}}{a^{2p}} \times \frac{a^{r+p}}{a^{2q}}$
$= a^{p+q-2r} \cdot a^{q+r-2p} \cdot a^{r+p-2q}$
$= a^{(p+q-2r) + (q+r-2p) + (r+p-2q)}$
$= a^{2p + 2q + 2r – 2p – 2q – 2r}$
$= a^0 = 1 = \text{R.H.S.}$
$\therefore \frac{a^{p+q}}{a^{2r}} \times \frac{a^{q+r}}{a^{2p}} \times \frac{a^{r+p}}{a^{2q}} = 1$ (Proved)
13. $\left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}} \cdot \left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}} \cdot \left(\frac{x^c}{x^a}\right)^{\frac{1}{ca}} = 1$
Solution:
$\text{L.H.S.} = \left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}} \cdot \left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}} \cdot \left(\frac{x^c}{x^a}\right)^{\frac{1}{ca}}$
$= \left(x^{a-b}\right)^{\frac{1}{ab}} \cdot \left(x^{b-c}\right)^{\frac{1}{bc}} \cdot \left(x^{c-a}\right)^{\frac{1}{ca}}$
$= x^{\frac{a-b}{ab}} \cdot x^{\frac{b-c}{bc}} \cdot x^{\frac{c-a}{ca}}$
$= x^{\frac{a-b}{ab} + \frac{b-c}{bc} + \frac{c-a}{ca}}$
$= x^{\frac{c(a-b) + a(b-c) + b(c-a)}{abc}}$
$= x^{\frac{ca – bc + ab – ca + bc – ab}{abc}}$
$= x^{\frac{0}{abc}} = x^0 = 1 = \text{R.H.S.}$
$\therefore \left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}} \cdot \left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}} \cdot \left(\frac{x^c}{x^a}\right)^{\frac{1}{ca}} = 1$ (Proved)
14. $\left(\frac{x^a}{x^b}\right)^{a+b} \cdot \left(\frac{x^b}{x^c}\right)^{b+c} \cdot \left(\frac{x^c}{x^a}\right)^{c+a} = 1$
Solution:
$\text{L.H.S.} = \left(\frac{x^a}{x^b}\right)^{a+b} \cdot \left(\frac{x^b}{x^c}\right)^{b+c} \cdot \left(\frac{x^c}{x^a}\right)^{c+a}$
$= (x^{a-b})^{a+b} \cdot (x^{b-c})^{b+c} \cdot (x^{c-a})^{c+a}$
$= x^{(a-b)(a+b)} \cdot x^{(b-c)(b+c)} \cdot x^{(c-a)(c+a)}$
$= x^{a^2 – b^2} \cdot x^{b^2 – c^2} \cdot x^{c^2 – a^2}$
$= x^{a^2 – b^2 + b^2 – c^2 + c^2 – a^2}$
$= x^0 = 1 = \text{R.H.S.}$
$\therefore \left(\frac{x^a}{x^b}\right)^{a+b} \cdot \left(\frac{x^b}{x^c}\right)^{b+c} \cdot \left(\frac{x^c}{x^a}\right)^{c+a} = 1$ (Proved)
15. $\left(\frac{x^p}{x^q}\right)^{p+q-r} \cdot \left(\frac{x^q}{x^r}\right)^{q+r-p} \cdot \left(\frac{x^r}{x^p}\right)^{r+p-q} = 1$
Solution:
$\text{L.H.S.} = \left(\frac{x^p}{x^q}\right)^{p+q-r} \cdot \left(\frac{x^q}{x^r}\right)^{q+r-p} \cdot \left(\frac{x^r}{x^p}\right)^{r+p-q}$
$= (x^{p-q})^{p+q-r} \cdot (x^{q-r})^{q+r-p} \cdot (x^{r-p})^{r+p-q}$
$= x^{(p-q)(p+q-r)} \cdot x^{(q-r)(q+r-p)} \cdot x^{(r-p)(r+p-q)}$
$= x^{p^2 – q^2 – pr + qr} \cdot x^{q^2 – r^2 – pq + pr} \cdot x^{r^2 – p^2 – qr + pq}$
$= x^{(p^2 – q^2 – pr + qr) + (q^2 – r^2 – pq + pr) + (r^2 – p^2 – qr + pq)}$
$= x^0 = 1 = \text{R.H.S.}$
$\therefore \left(\frac{x^p}{x^q}\right)^{p+q-r} \cdot \left(\frac{x^q}{x^r}\right)^{q+r-p} \cdot \left(\frac{x^r}{x^p}\right)^{r+p-q} = 1$ (Proved)
16. If $a^x = b$, $b^y = c$, and $c^z = a$, then show that $xyz = 1$.
Solution: Given, $a^x = b$, $b^y = c$, and $c^z = a$,
Now, $c^z = a$
or, $(b^y)^z = a \quad [\because c = b^y]$
or, $b^{yz} = a$
or, $(a^x)^{yz} = a \quad [\because b = a^x]$
or, $a^{xyz} = a^1$
$\therefore xyz = 1$ (Shown)
Solve (17 – 20):
17. $4^x = 8$
Solution: $4^x = 8$
or, $(2^2)^x = 2^3$
or, $2^{2x} = 2^3$
or, $2x = 3$
or, $x = \frac{3}{2}$
Answer: $x = \frac{3}{2}$
18. $2^{2x+1} = 128$
Solution: $2^{2x+1} = 128$
or, $2^{2x+1} = 2^7 \quad [\because 2^7 = 128]$
$2x + 1 = 7$
or, $2x = 7 – 1$
or, $2x = 6$
or, $x = \frac{6}{2}$
or, $x = 3$
Answer: $x = 3$
19. $(\sqrt{3})^{x+1} = (\sqrt[3]{3})^{2x-1}$
Solution: $(\sqrt{3})^{x+1} = (\sqrt[3]{3})^{2x-1}$
or, $(3^{\frac{1}{2}})^{x+1} = (3^{\frac{1}{3}})^{2x-1}$
or, $3^{\frac{x+1}{2}} = 3^{\frac{2x-1}{3}}$
or, $\frac{x+1}{2} = \frac{2x-1}{3}$
or, $2(2x – 1) = 3(x + 1)$
or, $4x – 2 = 3x + 3$
or, $4x – 3x = 3 + 2$
or, $x = 5$
Answer: $x = 5$
20. $2^x + 2^{1-x} = 3$
Solution: $2^x + 2^{1-x} = 3$
or, $2^x + \frac{2^1}{2^x} = 3$
Let $2^x = a$.
Then, $a + \frac{2}{a} = 3$
or, $\frac{a^2 + 2}{a} = 3$
or, $a^2 + 2 = 3a$
or, $a^2 – 3a + 2 = 0$
or, $a^2 – 2a – a + 2 = 0$
or, $a(a – 2) – 1(a – 2) = 0$
or, $(a – 2)(a – 1) = 0$
Either, $a – 2 = 0 \implies a = 2$
Or, $a – 1 = 0 \implies a = 1$
Now, substituting the value of $a$, we get
$2^x = 2^1 \implies x = 1$
and $2^x = 1 \implies 2^x = 2^0 \implies x = 0$
Answer: $x = 0, 1$