Exercise 9.1: Exponential Problems
1. Prove that, $\left(a^{\frac{m}{n}}\right)^p = a^{\frac{mp}{n}}$, where $m, p \in \mathbb{Z}$ and $n \in \mathbb{N}$.
Solution: Let, $x = a^{\frac{1}{n}}$
Then, according to the basic definition of exponents, $x^n = a$
$\text{L.H.S.} = \left(a^{\frac{m}{n}}\right)^p$
$= \left\{\left(a^{\frac{1}{n}}\right)^m\right\}^p$
$= (x^m)^p$
$= x^{mp}$
Substituting back $x = a^{\frac{1}{n}}$:
$= \left(a^{\frac{1}{n}}\right)^{mp}$
$= a^{\frac{mp}{n}} = \text{R.H.S.}$
(Proved)
2. Prove that, $\left(a^{\frac{1}{m}}\right)^{\frac{1}{n}} = a^{\frac{1}{mn}}$, where $m, n \in \mathbb{Z}, m \neq 0, n \neq 0$.
Solution: Let, $x = \left(a^{\frac{1}{m}}\right)^{\frac{1}{n}}$
or, $x^n = \left\{\left(a^{\frac{1}{m}}\right)^{\frac{1}{n}}\right\}^n$ [Taking the $n$-th power on both sides]
or, $x^n = a^{\frac{1}{m}}$
or, $(x^n)^m = \left(a^{\frac{1}{m}}\right)^m$ [Taking the $m$-th power on both sides again]
or, $x^{mn} = a$
or, $x = a^{\frac{1}{mn}}$ [Taking the $\frac{1}{mn}$-th power on both sides]
Therefore, $\left(a^{\frac{1}{m}}\right)^{\frac{1}{n}} = a^{\frac{1}{mn}}$
(Proved)
3. Prove that, $(ab)^{\frac{m}{n}} = (a)^{\frac{m}{n}}(b)^{\frac{m}{n}}$, where $m \in \mathbb{Z}, n \in \mathbb{N}$.
Solution: Let, $x = a^{\frac{1}{n}}$ and $y = b^{\frac{1}{n}}$
According to the rules of exponents, $x^n = a$ and $y^n = b$
Now,
$ab = x^n \cdot y^n = (xy)^n$
Taking the $\frac{1}{n}$-th power on both sides:
$(ab)^{\frac{1}{n}} = xy = a^{\frac{1}{n}} \cdot b^{\frac{1}{n}}$
Raising both sides to the power $m$, we get
$\left\{(ab)^{\frac{1}{n}}\right\}^m = \left\{a^{\frac{1}{n}} \cdot b^{\frac{1}{n}}\right\}^m$
or, $(ab)^{\frac{m}{n}} = \left(a^{\frac{1}{n}}\right)^m \cdot \left(b^{\frac{1}{n}}\right)^m$
or, $(ab)^{\frac{m}{n}} = (a)^{\frac{m}{n}}(b)^{\frac{m}{n}}$
(Proved)
4. Show that,
1) $\left(a^{\frac{1}{3}} – b^{\frac{1}{3}}\right)\left(a^{\frac{2}{3}} + a^{\frac{1}{3}}b^{\frac{1}{3}} + b^{\frac{2}{3}}\right) = a – b$
Solution: Let, $x = a^{\frac{1}{3}}$ and $y = b^{\frac{1}{3}}$
Then, $x^2 = a^{\frac{2}{3}}$ and $y^2 = b^{\frac{2}{3}}$
$\text{L.H.S.} = \left(a^{\frac{1}{3}} – b^{\frac{1}{3}}\right)\left(a^{\frac{2}{3}} + a^{\frac{1}{3}}b^{\frac{1}{3}} + b^{\frac{2}{3}}\right)$
$= (x – y)(x^2 + xy + y^2)$
$= x^3 – y^3$
$= \left(a^{\frac{1}{3}}\right)^3 – \left(b^{\frac{1}{3}}\right)^3$
$= a^{3 \times \frac{1}{3}} – b^{3 \times \frac{1}{3}}$
$= a – b = \text{R.H.S.}$
(Shown)
2) $\frac{a^3 + a^{-3} + 1}{a^{\frac{3}{2}} + a^{-\frac{3}{2}} + 1} = a^{\frac{3}{2}} + a^{-\frac{3}{2}} – 1$
Solution: Let, $x = a^{\frac{3}{2}}$ and $x^{-1} = a^{-\frac{3}{2}}$
Then, $x^2 = \left(a^{\frac{3}{2}}\right)^2 = a^3$ and $x^{-2} = \left(a^{-\frac{3}{2}}\right)^2 = a^{-3}$
$\text{L.H.S.} = \frac{a^3 + a^{-3} + 1}{a^{\frac{3}{2}} + a^{-\frac{3}{2}} + 1}$
$= \frac{x^2 + x^{-2} + 1}{x + x^{-1} + 1}$
$= \frac{x^2 + 2 \cdot x \cdot x^{-1} + x^{-2} – 1}{x + x^{-1} + 1}$ [Since $x \cdot x^{-1} = 1$]
$= \frac{(x + x^{-1})^2 – (1)^2}{x + x^{-1} + 1}$
$= \frac{(x + x^{-1} + 1)(x + x^{-1} – 1)}{x + x^{-1} + 1}$
$= x + x^{-1} – 1$
$= a^{\frac{3}{2}} + a^{-\frac{3}{2}} – 1$ [Substituting $x = a^{\frac{3}{2}}$ and $x^{-1} = a^{-\frac{3}{2}}$]
$= \text{R.H.S.}$
(Shown)
5. Simplify:
1) $\frac{\left(\frac{a + b}{b}\right)^{\frac{a}{a – b}} \times \left(\frac{a – b}{a}\right)^{\frac{a}{a – b}}}{\left(\frac{a + b}{b}\right)^{\frac{b}{a – b}} \times \left(\frac{a – b}{a}\right)^{\frac{b}{a – b}}}$
Solution:
$\frac{\left(\frac{a + b}{b}\right)^{\frac{a}{a – b}} \times \left(\frac{a – b}{a}\right)^{\frac{a}{a – b}}}{\left(\frac{a + b}{b}\right)^{\frac{b}{a – b}} \times \left(\frac{a – b}{a}\right)^{\frac{b}{a – b}}}$
$= \frac{\left[\frac{a+b}{b} \times \frac{a-b}{a}\right]^{\frac{a}{a-b}}}{\left[\frac{a+b}{b} \times \frac{a-b}{a}\right]^{\frac{b}{a-b}}}$
$= \left[\frac{(a+b)(a-b)}{ab}\right]^{\frac{a}{a-b} – \frac{b}{a-b}}$
$= \left[\frac{a^2 – b^2}{ab}\right]^{\frac{a-b}{a-b}}$
$= \left[\frac{a^2 – b^2}{ab}\right]^1$
$= \frac{a^2 – b^2}{ab}$
Answer: $\frac{a^2 – b^2}{ab}$
2) $\frac{a^{\frac{3}{2}} + ab}{ab – b^3} – \frac{\sqrt{a}}{\sqrt{a} – b}$
Solution:
$\frac{a^{\frac{3}{2}} + ab}{ab – b^3} – \frac{\sqrt{a}}{\sqrt{a} – b}$
$= \frac{a \cdot a^{\frac{1}{2}} + ab}{b(a – b^2)} – \frac{\sqrt{a}}{\sqrt{a} – b}$
$= \frac{a\sqrt{a} + ab}{b(\sqrt{a} – b)(\sqrt{a} + b)} – \frac{\sqrt{a}}{\sqrt{a} – b}$
$= \frac{a(\sqrt{a} + b)}{b(\sqrt{a} – b)(\sqrt{a} + b)} – \frac{\sqrt{a}}{\sqrt{a} – b}$
$= \frac{a}{b(\sqrt{a} – b)} – \frac{\sqrt{a}}{\sqrt{a} – b}$
$= \frac{a – b\sqrt{a}}{b(\sqrt{a} – b)}$
$= \frac{\sqrt{a}(\sqrt{a} – b)}{b(\sqrt{a} – b)}$
$= \frac{\sqrt{a}}{b}$
Answer: $\frac{\sqrt{a}}{b}$
3) $\frac{1}{1 + a^{-m}b^n + a^{-m}c^p} + \frac{1}{1 + b^{-n}c^p + b^{-n}a^m} + \frac{1}{1 + c^{-p}a^m + c^{-p}b^n}$
Solution:
$\frac{1}{1 + a^{-m}b^n + a^{-m}c^p} + \frac{1}{1 + b^{-n}c^p + b^{-n}a^m} + \frac{1}{1 + c^{-p}a^m + c^{-p}b^n}$
$= \frac{1}{1 + \frac{b^n}{a^m} + \frac{c^p}{a^m}} + \frac{1}{1 + \frac{c^p}{b^n} + \frac{a^m}{b^n}} + \frac{1}{1 + \frac{a^m}{c^p} + \frac{b^n}{c^p}}$
$= \frac{1}{\frac{a^m + b^n + c^p}{a^m}} + \frac{1}{\frac{b^n + c^p + a^m}{b^n}} + \frac{1}{\frac{c^p + a^m + b^n}{c^p}}$
$= \frac{a^m}{a^m + b^n + c^p} + \frac{b^n}{a^m + b^n + c^p} + \frac{c^p}{a^m + b^n + c^p}$
$= \frac{a^m + b^n + c^p}{a^m + b^n + c^p}$
$= 1$
Answer: $1$
4) $\sqrt[bc]{\frac{x^{\frac{b}{c}}}{x^{\frac{c}{b}}}} \times \sqrt[ca]{\frac{x^{\frac{c}{a}}}{x^{\frac{a}{c}}}} \times \sqrt[ab]{\frac{x^{\frac{a}{b}}}{x^{\frac{b}{a}}}}$
Solution:
$\sqrt[bc]{\frac{x^{\frac{b}{c}}}{x^{\frac{c}{b}}}} \times \sqrt[ca]{\frac{x^{\frac{c}{a}}}{x^{\frac{a}{c}}}} \times \sqrt[ab]{\frac{x^{\frac{a}{b}}}{x^{\frac{b}{a}}}}$
$= \left(x^{\frac{b}{c} – \frac{c}{b}}\right)^{\frac{1}{bc}} \times \left(x^{\frac{c}{a} – \frac{a}{c}}\right)^{\frac{1}{ca}} \times \left(x^{\frac{a}{b} – \frac{b}{a}}\right)^{\frac{1}{ab}}$
$= \left(x^{\frac{b^2 – c^2}{bc}}\right)^{\frac{1}{bc}} \times \left(x^{\frac{c^2 – a^2}{ca}}\right)^{\frac{1}{ca}} \times \left(x^{\frac{a^2 – b^2}{ab}}\right)^{\frac{1}{ab}}$
$= x^{\frac{b^2 – c^2}{b^2c^2}} \times x^{\frac{c^2 – a^2}{c^2a^2}} \times x^{\frac{a^2 – b^2}{a^2b^2}}$
$= x^{\frac{b^2 – c^2}{b^2c^2} + \frac{c^2 – a^2}{c^2a^2} + \frac{a^2 – b^2}{a^2b^2}}$
$= x^{\frac{a^2(b^2 – c^2) + b^2(c^2 – a^2) + c^2(a^2 – b^2)}{a^2b^2c^2}}$
$= x^{\frac{a^2b^2 – a^2c^2 + b^2c^2 – a^2b^2 + a^2c^2 – b^2c^2}{a^2b^2c^2}}$
$= x^{\frac{0}{a^2b^2c^2}}$
$= x^0$
$= 1$
Answer: $1$
5) $\frac{(a^2 – b^{-2})^a (a – b^{-1})^{b – a}}{(b^2 – a^{-2})^b (b + a^{-1})^{a – b}}$
Solution:
$\frac{(a^2 – b^{-2})^a (a – b^{-1})^{b – a}}{(b^2 – a^{-2})^b (b + a^{-1})^{a – b}}$
$= \frac{\left(a + \frac{1}{b}\right)^a \left(a – \frac{1}{b}\right)^a \left(a – \frac{1}{b}\right)^{b – a}}{\left(b + \frac{1}{a}\right)^b \left(b – \frac{1}{a}\right)^b \left(b + \frac{1}{a}\right)^{a – b}}$
$= \frac{\left(\frac{ab + 1}{b}\right)^a \left(\frac{ab – 1}{b}\right)^{a + b – a}}{\left(\frac{ab + 1}{a}\right)^{b + a – b} \left(\frac{ab – 1}{a}\right)^b}$
$= \frac{\left(\frac{ab + 1}{b}\right)^a \left(\frac{ab – 1}{b}\right)^b}{\left(\frac{ab + 1}{a}\right)^a \left(\frac{ab – 1}{a}\right)^b}$
$= \frac{\frac{(ab + 1)^a}{b^a} \cdot \frac{(ab – 1)^b}{b^b}}{\frac{(ab + 1)^a}{a^a} \cdot \frac{(ab – 1)^b}{a^b}}$
$= \frac{\frac{(ab + 1)^a (ab – 1)^b}{b^{a+b}}}{\frac{(ab + 1)^a (ab – 1)^b}{a^{a+b}}}$
$= \frac{a^{a+b}}{b^{a+b}}$
$= \left(\frac{a}{b}\right)^{a+b}$
Answer: $\left(\frac{a}{b}\right)^{a+b}$
6. Show that,
1) If $x = a^{q+r}b^p$, $y = a^{r+p}b^q$, $z = a^{p+q}b^r$, then $x^{q-r} \cdot y^{r-p} \cdot z^{p-q} = 1$
Solution: Given,
$x = a^{q+r}b^p$
$y = a^{r+p}b^q$
$z = a^{p+q}b^r$
$\text{L.H.S.} = x^{q-r} \cdot y^{r-p} \cdot z^{p-q}$
$= (a^{q+r}b^p)^{q-r} \cdot (a^{r+p}b^q)^{r-p} \cdot (a^{p+q}b^r)^{p-q}$
$= \left(a^{(q+r)(q-r)} \cdot b^{p(q-r)}\right) \cdot \left(a^{(r+p)(r-p)} \cdot b^{q(r-p)}\right) \cdot \left(a^{(p+q)(p-q)} \cdot b^{r(p-q)}\right)$
$= \left(a^{q^2 – r^2} \cdot b^{pq – pr}\right) \cdot \left(a^{r^2 – p^2} \cdot b^{qr – pq}\right) \cdot \left(a^{p^2 – q^2} \cdot b^{pr – qr}\right)$
$= a^{(q^2 – r^2) + (r^2 – p^2) + (p^2 – q^2)} \cdot b^{(pq – pr) + (qr – pq) + (pr – qr)}$
$= a^0 \cdot b^0$
$= 1 \times 1$
$= 1 = \text{R.H.S.}$
(Shown)
2) If $a^p = b$, $b^q = c$, and $c^r = a$, then $pqr = 1$
Solution: Given,
$c^r = a$
or, $(b^q)^r = a$ [Since $c = b^q$]
or, $b^{qr} = a$
or, $(a^p)^{qr} = a$ [Since $b = a^p$]
or, $a^{pqr} = a^1$
or, $pqr = 1 = \text{R.H.S.}$
(Shown)
3) If $a^x = p$, $a^y = q$, and $a^2 = (p^y q^x)^z$, then $xyz = 1$
Solution: Given,
$a^2 = (p^y q^x)^z$
Substituting $p = a^x$ and $q = a^y$:
$a^2 = \left((a^x)^y \cdot (a^y)^x\right)^z$
or, $a^2 = (a^{xy} \cdot a^{xy})^z$
or, $a^2 = (a^{xy + xy})^z$
or, $a^2 = (a^{2xy})^z$
or, $a^2 = a^{2xyz}$
or, $2 = 2xyz$
or, $2xyz = 2$
or, $xyz = \frac{2}{2}$
or, $xyz = 1 = \text{R.H.S.}$
(Shown)
7.
1) If $x\sqrt[3]{a} + y\sqrt[3]{b} + z\sqrt[3]{c} = 0$ and $a^2 = bc$, then show that $ax^3 + by^3 + cz^3 = 3axyz$
Solution: Given, $x\sqrt[3]{a} + y\sqrt[3]{b} + z\sqrt[3]{c} = 0$
or, $x\sqrt[3]{a} + y\sqrt[3]{b} = -z\sqrt[3]{c}$
or, $(x\sqrt[3]{a} + y\sqrt[3]{b})^3 = (-z\sqrt[3]{c})^3$ [Cubing both sides]
or, $(x\sqrt[3]{a})^3 + (y\sqrt[3]{b})^3 + 3(x\sqrt[3]{a})(y\sqrt[3]{b})(x\sqrt[3]{a} + y\sqrt[3]{b}) = -z^3 c$
or, $x^3 a + y^3 b + 3xy\sqrt[3]{ab}(-z\sqrt[3]{c}) = -cz^3$
or, $ax^3 + by^3 – 3xyz\sqrt[3]{abc} = -cz^3$
or, $ax^3 + by^3 – 3xyz\sqrt[3]{a^3} = -cz^3$ [Since $a^2 = bc$, we have $abc = a \cdot a^2 = a^3$]
or, $ax^3 + by^3 – 3xyz(a) = -cz^3$
or, $ax^3 + by^3 + cz^3 = 3axyz$
(Shown)
2) If $x = (a + b)^{\frac{1}{3}} + (a – b)^{\frac{1}{3}}$ and $a^2 – b^2 = c^3$, show that $x^3 – 3cx – 2a = 0$
Solution: Given, $x = (a + b)^{\frac{1}{3}} + (a – b)^{\frac{1}{3}}$
or, $x^3 = \left\{(a + b)^{\frac{1}{3}} + (a – b)^{\frac{1}{3}}\right\}^3$ [Cubing both sides]
or, $x^3 = \left((a + b)^{\frac{1}{3}}\right)^3 + \left((a – b)^{\frac{1}{3}}\right)^3 + 3(a + b)^{\frac{1}{3}}(a – b)^{\frac{1}{3}}\left\{(a + b)^{\frac{1}{3}} + (a – b)^{\frac{1}{3}}\right\}$
or, $x^3 = (a + b) + (a – b) + 3\{(a + b)(a – b)\}^{\frac{1}{3}} \cdot x$
or, $x^3 = 2a + 3(a^2 – b^2)^{\frac{1}{3}} \cdot x$
or, $x^3 = 2a + 3(c^3)^{\frac{1}{3}} \cdot x$ [Since $a^2 – b^2 = c^3$]
or, $x^3 = 2a + 3cx$
or, $x^3 – 3cx – 2a = 0$
(Shown)
3) If $a^2 + 2 = 3^{\frac{2}{3}} + 3^{-\frac{2}{3}}$ and $a \geq 0$, show that $3a^3 + 9a = 8$
Solution: Given, $a^2 + 2 = 3^{\frac{2}{3}} + 3^{-\frac{2}{3}}$
or, $a^2 = 3^{\frac{2}{3}} + 3^{-\frac{2}{3}} – 2$
or, $a^2 = \left(3^{\frac{1}{3}}\right)^2 – 2 \cdot 3^{\frac{1}{3}} \cdot 3^{-\frac{1}{3}} + \left(3^{-\frac{1}{3}}\right)^2$
or, $a^2 = \left(3^{\frac{1}{3}} – 3^{-\frac{1}{3}}\right)^2$
$a = 3^{\frac{1}{3}} – 3^{-\frac{1}{3}}$ [Taking square root on both sides (since $a \geq 0$)]
$a^3 = \left(3^{\frac{1}{3}} – 3^{-\frac{1}{3}}\right)^3$ [Cubing both sides]
or, $a^3 = \left(3^{\frac{1}{3}}\right)^3 – \left(3^{-\frac{1}{3}}\right)^3 – 3 \cdot 3^{\frac{1}{3}} \cdot 3^{-\frac{1}{3}} \left(3^{\frac{1}{3}} – 3^{-\frac{1}{3}}\right)$
or, $a^3 = 3 – 3^{-1} – 3(1)(a)$
or, $a^3 = 3 – \frac{1}{3} – 3a$
or, $a^3 + 3a = \frac{8}{3}$
Multiplying both sides by $3$, we get
$3a^3 + 9a = 8$
(Shown)
4) If $a^2 = b^3$, then show that $\left(\frac{a}{b}\right)^{\frac{3}{2}} + \left(\frac{b}{a}\right)^{\frac{2}{3}} = a^{\frac{1}{2}} + b^{-\frac{1}{3}}$
Solution: Given, $a^2 = b^3 \implies a = b^{\frac{3}{2}}$ and $b = a^{\frac{2}{3}}$
$\text{L.H.S.} = \left(\frac{a}{b}\right)^{\frac{3}{2}} + \left(\frac{b}{a}\right)^{\frac{2}{3}}$
$= \frac{a^{\frac{3}{2}}}{b^{\frac{3}{2}}} + \frac{b^{\frac{2}{3}}}{a^{\frac{2}{3}}}$
$= \frac{a^{\frac{3}{2}}}{a} + \frac{b^{\frac{2}{3}}}{b}$ [Substituting $b^{\frac{3}{2}} = a$ and $a^{\frac{2}{3}} = b$]
$= a^{\frac{3}{2} – 1} + b^{\frac{2}{3} – 1}$
$= a^{\frac{1}{2}} + b^{-\frac{1}{3}} = \text{R.H.S.}$
(Shown)
5) If $b = 1 + 3^{\frac{2}{3}} + 3^{\frac{1}{3}}$, then show that $b^3 – 3b^2 – 6b – 4 = 0$
Solution: Given, $b = 1 + 3^{\frac{2}{3}} + 3^{\frac{1}{3}}$
or, $b – 1 = 3^{\frac{2}{3}} + 3^{\frac{1}{3}}$
or, $(b – 1)^3 = \left(3^{\frac{2}{3}} + 3^{\frac{1}{3}}\right)^3$ [Cubing both sides]
or, $b^3 – 3b^2 + 3b – 1 = \left(3^{\frac{2}{3}}\right)^3 + \left(3^{\frac{1}{3}}\right)^3 + 3 \cdot 3^{\frac{2}{3}} \cdot 3^{\frac{1}{3}}\left(3^{\frac{2}{3}} + 3^{\frac{1}{3}}\right)$
or, $b^3 – 3b^2 + 3b – 1 = 3^2 + 3^1 + 3 \cdot 3^{1}(b – 1)$
or, $b^3 – 3b^2 + 3b – 1 = 9 + 3 + 9(b – 1)$
or, $b^3 – 3b^2 + 3b – 1 = 12 + 9b – 9$
or, $b^3 – 3b^2 + 3b – 1 = 9b + 3$
or, $b^3 – 3b^2 + 3b – 9b – 1 – 3 = 0$
or, $b^3 – 3b^2 – 6b – 4 = 0$
(Shown)
6) If $a + b + c = 0$, show that $\frac{1}{x^b + x^{-c} + 1} + \frac{1}{x^c + x^{-a} + 1} + \frac{1}{x^a + x^{-b} + 1} = 1$
Solution: Given, $a + b + c = 0$
Therefore, $-c = a + b$, $-a = b + c$, and $-b = c + a$
1st term $= \frac{1}{x^b + x^{-c} + 1} = \frac{1}{x^b + x^{a+b} + 1}$
2nd term $= \frac{1}{x^c + x^{-a} + 1} = \frac{1}{x^c + x^{b+c} + 1} = \frac{x^{-c}}{1 + x^b + x^{-c}} = \frac{x^{-c}}{1 + x^b + x^{a+b}}$
3rd term $= \frac{1}{x^a + x^{-b} + 1} = \frac{x^{-a}}{1 + x^{-a-b} + x^{-a}} = \frac{x^{b}}{x^{a+b} + 1 + x^b}$
$\text{L.H.S.} = \frac{1}{x^b + x^{-c} + 1} + \frac{1}{x^c + x^{-a} + 1} + \frac{1}{x^a + x^{-b} + 1}$
$= \frac{1}{x^a + x^{-b} + 1} + \frac{x^a}{x^a + x^{-b} + 1} + \frac{x^{-b}}{x^a + x^{-b} + 1}$
$= \frac{1 + x^a + x^{-b}}{1 + x^a + x^{-b}}$
$= 1 = \text{R.H.S.}$
(Shown)
8.
1) If $a^x = b$, $b^y = c$, and $c^z = 1$, then find the value of $xyz$.
Solution: Given, $c^z = 1$
or, $(b^y)^z = 1$
or, $b^{yz} = 1$
or, $(a^x)^{yz} = 1$
or, $a^{xyz} = 1$
or, $a^{xyz} = a^0$
or, $xyz = 0$
Answer: $0$
2) If $x^a = y^b = z^c$ and $xyz = 1$, then find the value of $ab + bc + ca$.
Solution: Let, $x^a = y^b = z^c = k$
Then, $x = k^{\frac{1}{a}}$, $y = k^{\frac{1}{b}}$, $z = k^{\frac{1}{c}}$
Given, $xyz = 1$
or, $k^{\frac{1}{a}} \cdot k^{\frac{1}{b}} \cdot k^{\frac{1}{c}} = 1$
or, $k^{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} = k^0$
or, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$
or, $\frac{bc + ca + ab}{abc} = 0$
or, $ab + bc + ca = 0$
Answer: $0$
3) If $9^x = 27^y$, then find the value of $\frac{x}{y}$.
Solution: Given, $9^x = 27^y$
or, $(3^2)^x = (3^3)^y$
or, $3^{2x} = 3^{3y}$
or, $2x = 3y$
or, $\frac{x}{y} = \frac{3}{2}$
Answer: $\frac{3}{2}$
9. Solve:
1) $3^{2x+2} + 27^{x+1} = 36$
Solution: $3^{2x+2} + 27^{x+1} = 36$
or, $3^{2x+2} + (3^3)^{x+1} = 36$
or, $3^{2x} \cdot 3^2 + 3^{3x+3} = 36$
or, $9 \cdot 3^{2x} + 27 \cdot 3^{3x} = 36$
Let $3^x = u$, we get
or, $9u^2 + 27u^3 = 36$
or, $27u^3 + 9u^2 – 36 = 0$
or, $3u^3 + u^2 – 4 = 0$ [Dividing both sides by 9]
or, $3u^3 – 3u^2 + 4u^2 – 4u + 4u – 4 = 0$
or, $3u^2(u – 1) + 4u(u – 1) + 4(u – 1) = 0$
or, $(u – 1)(3u^2 + 4u + 4) = 0$
Here, $3u^2 + 4u + 4 = 0$ has no real roots (discriminant $16 – 48 < 0$).
Therefore, $u – 1 = 0$
or, $u = 1$
or, $3^x = 1$ [Substituting $u = 3^x$]
or, $3^x = 3^0$
or, $x = 0$
Answer: $x = 0$
2) $5^x + 3^y = 8, 5^{x-1} + 3^{y-1} = 2$
Solution: Let $5^x = u$ and $3^y = v$
1st equation: $u + v = 8 \quad \dots (1)$
2nd equation: $\frac{u}{5} + \frac{v}{3} = 2 \implies 3u + 5v = 30 \quad \dots (2)$
Multiplying equation (1) by $3$ and subtracting it from (2), we get
$(3u + 5v) – (3u + 3v) = 30 – 24$
or, $2v = 6$
or, $v = 3$
Substituting $v = 3$ in equation (1), we get
$u + 3 = 8 \implies u = 5$
Now,
$5^x = u = 5^1 \implies x = 1$
$3^y = v = 3^1 \implies y = 1$
Answer: $(x, y) = (1, 1)$
3) $4^{3y-2} = 16^{x+y}, 3^{x+2y} = 9^{2x+1}$
Solution: $4^{3y-2} = 16^{x+y}$
or, $4^{3y-2} = (4^2)^{x+y}$
or, $3y – 2 = 2x + 2y$
or, $2x – y = -2 \quad \dots (1)$
Also,
$3^{x+2y} = (3^2)^{2x+1}$
or, $x + 2y = 4x + 2$
or, $3x – 2y = -2 \quad \dots (2)$
Multiplying equation (1) by $2$ and subtracting (2) from it, we get
$(4x – 2y) – (3x – 2y) = -4 – (-2)$
or, $x = -2$
Substituting $x = -2$ in equation (1), we get
$2(-2) – y = -2$
or, $-4 – y = -2$
or, $y = -2$
Answer: $(x, y) = (-2, -2)$
(d) $2^{2x+1} \cdot 2^{3y+1} = 8, 2^{x+2} \cdot 2^{y+2} = 16$
Solution:
$2^{2x+1} \cdot 2^{3y+1} = 8$
or, $2^{(2x+1) + (3y+1)} = 2^3$
or, $2x + 3y + 2 = 3$
or, $2x + 3y = 1 \quad \dots (1)$
Also,
$2^{x+2} \cdot 2^{y+2} = 16$
or, $2^{(x+2) + (y+2)} = 2^4$
or, $x + y + 4 = 4$
or, $x + y = 0 \implies x = -y \quad \dots (2)$
Substituting $x = -y$ in equation (1), we get
$2(-y) + 3y = 1$
or, $y = 1$
Therefore, $x = -1$
Answer: $(x, y) = (-1, 1)$