Indices — Chapter Test

If 2^x.3^y.5^z = 2160, find the values of x, y and z. Hence, compute the value of 3^x.2^-y.5^-z.

Answer

2160 = 2⁴.3³.5¹.

∴ 2⁴.3³.5¹ = 2^x.3^y.5^z.

∴ x = 4, y = 3 and z = 1.

Substituting values of x, y and z in 3^x.2^-y.5^-z

$= 3^4 \times 2^{-3} \times 5^{-1} \\[1em] = 81 \times \dfrac{1}{2^3} \times \dfrac{1}{5} \\[1em] = 81 \times \dfrac{1}{8} \times \dfrac{1}{5} \\[1em] = \dfrac{81}{40} = 2\dfrac{1}{40}.$

Hence, x = 4, y = 3 and z = 1 and 3^x.2^-y.5^-z = $2\dfrac{1}{40}.$

If x = 2 and y = -3, find the values of

(i) x^x + y^y

(ii) x^y + y^x

Answer

(i) x^x + y^y = 2² + (-3)^-3

= 4 + $\Big(-\dfrac{1}{3}\Big)^3$

= 4 + $-\dfrac{1}{27}$

= 4 - $\dfrac{1}{27}$

= $\dfrac{108 - 1}{27} = \dfrac{107}{27} = 3\dfrac{26}{27}$.

Hence, x^x + y^y = $3\dfrac{26}{27}$.

(ii) x^y + y^x

$x^y + y^x = 2^{-3} + (-3)^{2} \\[1em] = \Big(\dfrac{1}{2}\Big)^3 + 9 \\[1em] = \dfrac{1}{8} + 9 \\[1em] = \dfrac{1 + 72}{8} \\[1em] = \dfrac{73}{8} = 9\dfrac{1}{8}.$

Hence, x^y + y^x = $9\dfrac{1}{8}$.

If p = x^{m + n}.y^l, q = x^{n + l}.y^m and r = x^{l + m}.yⁿ, prove that

p^{m - n}.q^{n - l}.r^{l - m} = 1

Answer

Substituting value of p, q and r in p^{m - n}.q^{n - l}.r^{l - m} = 1.

= (x^{m + n}.y^l)^{m - n}.(x^{n + l}.y^m)^{n - l}.(x^{l + m}.yⁿ)^{l - m}

= x^{(m + n)(m - n)}.y^{l(m - n)}.x^{(n + l)(n - l)}.y^{m(n - l)}.x^{(l + m)(l - m)}.y^{n(l - m)}

= x^{m2 - n2}.x^{n2 - l2}.x^{l2 - m2}.y^{lm - ln}.y^{mn - ml}.y^{nl - nm}

= x^{m2 - n2 + n2 - l2 + l2 - m2}.y^{lm - ln + mn - ml + nl - nm}

= x⁰.y⁰

= 1.1

= 1.

Hence, proved that, p^{m - n}.q^{n - l}.r^{l - m} = 1.

If x = a^{m + n}, y = a^{n + l} and z = a^{l + m}, prove that x^myⁿz^l = xⁿy^lz^m.

Answer

Substituting values of x, y and z in L.H.S. of x^myⁿz^l = xⁿy^lz^m we get,

x^myⁿz^l = (a^{m + n})^m.(a^{n + l})ⁿ.(a^{l + m})^l

= a^{m2 + mn}.a^{n2 + ln}.a^{lm + l2}

= a^{m2 + n2 + l2 + mn + lm + ln} .......(i)

Substituting values of x, y and z in R.H.S. of x^myⁿz^l = xⁿy^lz^m we get,

xⁿy^lz^m = (a^{m + n})ⁿ.(a^{n + l})^l.(a^{l + m})^m

= a^{mn + n2}.a^{nl + l2}.a^{lm + m2}

= a^{m2 + n2 + l2 + mn + lm + ln} .........(ii)

Since (i) = (ii),

Hence, proved that x^myⁿz^l = xⁿy^lz^m.

Show that $\dfrac{\Big(p + \dfrac{1}{q}\Big)^m \times \Big(p - \dfrac{1}{q}\Big)^n}{\Big(q + \dfrac{1}{p}\Big)^m \times \Big(q - \dfrac{1}{p}\Big)^n} = \Big(\dfrac{p}{q}\Big)^{m + n}$

Answer

Given,

$\Rightarrow \dfrac{\Big(p + \dfrac{1}{q}\Big)^m \times \Big(p - \dfrac{1}{q}\Big)^n}{\Big(q + \dfrac{1}{p}\Big)^m \times \Big(q - \dfrac{1}{p}\Big)^n} = \dfrac{\Big(\dfrac{pq + 1}{q}\Big)^m \times \Big(\dfrac{pq - 1}{q}\Big)^n}{\Big(\dfrac{pq + 1}{p}\Big)^m \times \Big(\dfrac{pq - 1}{p}\Big)^n} \\[1em] = \dfrac{\dfrac{(pq + 1)^m}{q^m} \times \dfrac{(pq - 1)^n}{q^n}}{\dfrac{(pq + 1)^m}{p^m} \times \dfrac{(pq - 1)^n}{p^n}} \\[1em] = \dfrac{(pq + 1)^m \times (pq - 1)^n \times p^m \times p^n}{(pq + 1)^m \times (pq - 1)^n \times q^m \times q^n} \\[1em] = \dfrac{p^m \times p^n}{q^m \times q^n} \\[1em] = \dfrac{p^{m + n}}{q^{m + n}} \\[1em] = \Big(\dfrac{p}{q}\Big)^{m + n}.$

Hence, proved that $\dfrac{\Big(p + \dfrac{1}{q}\Big)^m \times \Big(p - \dfrac{1}{q}\Big)^n}{\Big(q + \dfrac{1}{p}\Big)^m \times \Big(q - \dfrac{1}{p}\Big)^n} = \Big(\dfrac{p}{q}\Big)^{m + n}$.

If x is a positive real number and exponents are rational numbers, then simplify the following :

$\dfrac{(x^{(a + b)})^2(x^{(b + c)})^2(x^{(c + a)})^2}{(x^ax^bx^c)^4}$

Answer

Given,

$\Rightarrow \dfrac{(x^{(a + b)})^2(x^{(b + c)})^2(x^{(c + a)})^2}{(x^ax^bx^c)^4} = \dfrac{x^{2a + 2b}.x^{2b +2c}.x^{2c + 2a}}{x^{(a + b + c)4}} \\[1em] = \dfrac{x^{2a + 2b + 2b + 2c + 2c + 2a}}{x^{4a + 4b + 4c}} \\[1em] = \dfrac{x^{4a + 4b + 4c}}{x^{4a + 4b + 4c}} \\[1em] = 1.$

Hence, $\dfrac{(x^{(a + b)})^2(x^{(b + c)})^2(x^{(c + a)})^2}{(x^ax^bx^c)^4}$ = 1.

If x is a positive real number and exponents are rational numbers, then simplify the following :

$\Big(\dfrac{x^{a^2}}{x^{b^2}}\Big)^{\dfrac{1}{a + b}}\Big(\dfrac{x^{b^2}}{x^{c^2}}\Big)^{\dfrac{1}{b + c}}\Big(\dfrac{x^{c^2}}{x^{a^2}}\Big)^{\dfrac{1}{c + a}}$

Answer

Given,

$\Rightarrow \Big(\dfrac{x^{a^2}}{x^{b^2}}\Big)^{\dfrac{1}{a + b}}\Big(\dfrac{x^{b^2}}{x^{c^2}}\Big)^{\dfrac{1}{b + c}}\Big(\dfrac{x^{c^2}}{x^{a^2}}\Big)^{\dfrac{1}{c + a}} \\[1em] \Rightarrow (x^{a^2 - b^2})^{\dfrac{1}{a + b}}(x^{b^2 - c^2})^{\dfrac{1}{b + c}}(x^{c^2 - a^2})^{\dfrac{1}{c + a}} \\[1em] \Rightarrow (x)^{\dfrac{a^2 - b^2}{a + b}}.(x)^{\dfrac{b^2 - c^2}{b + c}}.(x)^{\dfrac{c^2 - a^2}{c + a}} \\[1em] \Rightarrow (x)^{\dfrac{(a - b)(a + b)}{a + b}}.(x)^{\dfrac{(b - c)(b + c)}{b + c}}.(x)^{\dfrac{(c - a)(c + a)}{c + a}} \\[1em] \Rightarrow (x)^{a - b}.(x)^{b - c}.(x)^{c - a} \\[1em] \Rightarrow x^{a - b + b - c + c - a} \\[1em] \Rightarrow x^{0} = 1.$

Hence, $\Big(\dfrac{x^{a^2}}{x^{b^2}}\Big)^{\dfrac{1}{a + b}}\Big(\dfrac{x^{b^2}}{x^{c^2}}\Big)^{\dfrac{1}{b + c}}\Big(\dfrac{x^{c^2}}{x^{a^2}}\Big)^{\dfrac{1}{c + a}}$ = 1.

If x is a positive real number and exponents are rational numbers, then simplify the following :

$\Big(\dfrac{x^b}{x^c}\Big)^{b + c - a}\Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c}$

Answer

Given,

$\Rightarrow \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a}\Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \\[1em] \Rightarrow (x^{b - c})^{b + c - a}.(x^{c - a})^{c + a - b}.(x^{a - b})^{a + b - c} \\[1em] \Rightarrow x^{(b - c)(b + c - a)}.(x)^{(c - a)(c + a - b)}.(x)^{(a - b)(a + b - c)} \\[1em] \Rightarrow x^{b^2 + bc - ba - cb - c^2 + ca}.(x)^{c^2 + ca - cb - ac - a^2 + ab}.(x)^{a^2 + ab - ac - ba - b^2 + bc} \\[1em] \Rightarrow x^{b^2 - b^2 + bc - cb - cb + bc - ba + ab + ab - ba - c^2 + c^2 + ca + ca - ac - ac - a^2 + a^2} \\[1em] \Rightarrow x^0 = 1.$

Hence, $\Big(\dfrac{x^b}{x^c}\Big)^{b + c - a}\Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c}$ = 1.

Show that $\dfrac{1}{1 + a^{y - x} + a^{z - x}} + \dfrac{1}{1 + a^{z - y} + a^{x - y}} + \dfrac{1}{1 + a^{x - z} + a^{y - z}}$ = 1.

Answer

Given,

$\dfrac{1}{1 + a^{y - x} + a^{z - x}} + \dfrac{1}{1 + a^{z - y} + a^{x - y}} + \dfrac{1}{1 + a^{x - z} + a^{y - z}}$ = 1.

Solving L.H.S. of the above equation,

$\Rightarrow \dfrac{1}{1 + a^{y - x} + a^{z - x}} + \dfrac{1}{1 + a^{z - y} + a^{x - y}} + \dfrac{1}{1 + a^{x - z} + a^{y - z}} \\[1em] \Rightarrow \dfrac{1}{1 + a^ya^{-x} + a^{z}.a^{-x}} + \dfrac{1}{1 + a^z.a^{-y} + a^x.a^{-y}} + \dfrac{1}{1 + a^x.a^{-z} + a^y.a^{-z}} \\[1em] \Rightarrow \dfrac{1}{1 + \dfrac{a^y}{a^x} + \dfrac{a^z}{a^x}} + \dfrac{1}{1 + \dfrac{a^z}{a^y} + \dfrac{a^x}{a^y}} + \dfrac{1}{1 + \dfrac{a^x}{a^z} + \dfrac{a^y}{a^z}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{a^x + a^y + a^z}{a^x}} + \dfrac{1}{\dfrac{a^y + a^z + a^x}{a^y}} + \dfrac{1}{\dfrac{a^z + a^x + a^y}{a^z}} \\[1em] \Rightarrow \dfrac{a^x}{a^x + a^y + a^z} + \dfrac{a^y}{a^x + a^y + a^z} + \dfrac{a^z}{a^x + a^y + a^z} \\[1em] \Rightarrow \dfrac{a^x + a^y + a^z}{a^x + a^y + a^z} \\[1em] \Rightarrow 1.$

Hence, proved that,

$\dfrac{1}{1 + a^{y - x} + a^{z - x}} + \dfrac{1}{1 + a^{z - y} + a^{x - y}} + \dfrac{1}{1 + a^{x - z} + a^{y - z}}$ = 1.

If (3)^x = (5)^y = (75)^z, show that $z = \dfrac{xy}{2x + y}$

Answer

Let (3)^x = (5)^y = (75)^z = k

∴ (3)^x = k

⇒ 3 = $k^{\dfrac{1}{x}}$ ......(i)

∴ (5)^y = k

⇒ 5 = $k^{\dfrac{1}{y}}$ .......(ii)

∴ (75)^z = k

⇒ 75 = $k^{\dfrac{1}{z}}$

⇒ 3.5² = $k^{\dfrac{1}{z}}$

Substituting value of x and y from (i) and (ii) in above equation we get,

$\Rightarrow k^{\dfrac{1}{x}} \times (k^{\dfrac{1}{y}})^2 = k^{\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{1}{x}} \times k^{\dfrac{2}{y}} = k^{\dfrac{1}{z}} \\[1em] \Rightarrow k^{\dfrac{1}{x} + \dfrac{2}{y}} = k^{\dfrac{1}{z}} \\[1em] \Rightarrow \dfrac{1}{x} + \dfrac{2}{y} = \dfrac{1}{z} \\[1em] \Rightarrow \dfrac{y + 2x}{xy} = \dfrac{1}{z} \\[1em] \Rightarrow z = \dfrac{xy}{2x + y}.$

Hence, proved that $z = \dfrac{xy}{2x + y}$.

Solve the following equations for x :

$\sqrt{3^{x + 1}} = 27$

Answer

(i) Solving,

$\Rightarrow \sqrt{3^{x + 1}} = 27 \\[1em] \Rightarrow \sqrt{3^{x + 1}} = 3^3 \\[1em] \Rightarrow \sqrt{3^{x + 1}} = \sqrt{3^6} \\[1em] \Rightarrow x + 1 = 6 \\[1em] \Rightarrow x = 6 - 1 = 5.$

Hence, x = 5.

Solve the following equation :

$4^{2x} = (\sqrt[3]{16})^{-\frac{6}{y}} = (\sqrt{8})^2$

Answer

Given,

$\Rightarrow 4^{2x} = (\sqrt[3]{16})^{-\frac{6}{y}} = (\sqrt{8})^2 \\[1em] \therefore 4^{2x} = (\sqrt{8})^2 \\[1em] \Rightarrow (2^2)^{2x} = 8 \\[1em] \Rightarrow (2)^{4x} = (2^3) \\[1em] \Rightarrow 4x = 3 \\[1em] \Rightarrow x = \dfrac{3}{4}.$

Similarly,

$\Rightarrow (\sqrt[3]{16})^{-\frac{6}{y}} = (\sqrt{8})^2 \\[1em] \Rightarrow (16)^{\frac{1}{3} \times -\frac{6}{y}} = 8 \\[1em] \Rightarrow (16)^{-\frac{2}{y}} = 8 \\[1em] \Rightarrow (2^4)^{-\frac{2}{y}} = 2^3 \\[1em] \Rightarrow (2)^{-\frac{8}{y}} = 2^3 \\[1em] \Rightarrow -\dfrac{8}{y} = 3 \\[1em] \Rightarrow y = -\dfrac{8}{3}.$

Hence, x = $\dfrac{3}{4} \text{ and } y = -\dfrac{8}{3}$.

Solve the following equation :

3^{x - 1} × 5^{2y - 3} = 225

Answer

Given,

⇒ 3^{x - 1} × 5^{2y - 3} = 225

⇒ 3^{x - 1} × 5^{2y - 3} = 3².5²

∴ x - 1 = 2 and 2y - 3 = 2

⇒ x = 2 + 1 and 2y = 2 + 3

⇒ x = 3 and y = $\dfrac{5}{2}$.

Hence, x = 3 and y = $\dfrac{5}{2}$.

Solve the following equation :

8^{x + 1} = 16^{y + 2}, $\Big(\dfrac{1}{2}\Big)^{3 + x} = \Big(\dfrac{1}{4}\Big)^{3y}$

Answer

Given,

⇒ 8^{x + 1} = 16^{y + 2}

⇒ (2³)^{x + 1} = (2⁴)^{y + 2}

⇒ 2^{3x + 3} = 2^{4y + 8}

⇒ 3x + 3 = 4y + 8

⇒ 3x - 4y = 5 .......(i)

Given,

$\Rightarrow \Big(\dfrac{1}{2}\Big)^{3 + x} = \Big(\dfrac{1}{4}\Big)^{3y} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^{3 + x} = \Big[\Big(\dfrac{1}{2}\Big)^2\Big]^{3y} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^{3 + x} = \Big(\dfrac{1}{2}\Big)^{6y} \\[1em] \Rightarrow 3 + x = 6y \\[1em] \Rightarrow 6y - x = 3 .......(ii)$

Multiplying (ii) by 3 we get,

⇒ 18y - 3x = 9 .......(iii)

Adding (i) and (iii) we get,

⇒ 3x - 4y + (18y - 3x) = 5 + 9
⇒ 14y = 14
⇒ y = 1.

Substituting value of y in (i) we get,

⇒ 3x - 4(1) = 5
⇒ 3x - 4 = 5
⇒ 3x = 9
⇒ x = 3.

Hence, x = 3 and y = 1.