# Solve the following equations

Question:

Solve the following equations

(i) $3^{x+1}=27 \times 3^{4}$

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

(iii) $3^{x-1} \times 5^{2 y-3}=225$

(iv) $8^{x+1}=16^{y+2}$ and $(1 / 2)^{3+x}=(1 / 4)^{3 y}$

Solution:

(i) $3^{x+1}=27 \times 3^{4}$

$3^{x+1}=3^{3} \times 3^{4}$

$3^{x+1}=3^{3+4}$

x + 1 = 3 + 4 [By equating exponents]

x + 1 = 7

x = 7 − 1

x = 6

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

$\left(2^{2}\right)^{2 x}=\left(16^{\frac{1}{3}}\right)^{-\frac{6}{y}}=(\sqrt{8})^{2}$

$2^{4 x}=\left[\left(2^{4}\right)^{\frac{1}{3}}\right]^{-\frac{6}{y}}=\left(2^{\frac{3}{2}}\right)^{2}$

$2^{4 x}=\left(2^{\frac{4}{3}}\right)^{-\frac{6}{y}}=\left(2^{\frac{3}{2}}\right)^{2}$

$2^{4 x}=\left(2^{\frac{4}{3}}\right)^{-\frac{6}{y}}=2^{3}$

$2^{4 x}=2^{3}$

$4 x=3$ (By equating exponents)

$x=\frac{3}{4}$

$2^{-\frac{8}{y}}=2^{3}$

$-\frac{8}{y}=3$ (Byequatingexponents)

$y=\frac{-8}{3}$

(iii) $3^{x-1} \times 5^{2 y-3}=225$

$3^{x-1} \times 5^{2 y-3}=3^{2} \times 5^{2}$

x − 1 = 2 [By equating exponents]

x = 3

$3^{x-1} \times 5^{2 y-3}=3^{2} \times 5^{2}$

2y − 3 = 2 [By equating exponents]

2y = 5

y = 5/2

(iv) $8^{x+1}=16^{y+2}$ and $(1 / 2)^{3+x}=(1 / 4)^{3 y}$

$\left(2^{3}\right)^{x+1}$ and $\left(2^{-1}\right)^{3+x}=\left(2^{-2}\right)^{3 y}$

3x + 3 = 4y + 8 and − 3 − x = −6y

3x + 3 = 4y + 8 and 3 + x = 6y

3x + 3 = 4y + 8 and y = (3+x)/6

3x + 3 = 4y + 8... eq1

$y=\frac{3+x}{6} \ldots .$ eq 2

Substitute eq2 in eq1

$3 x+3=4\left(\frac{3+x}{6}\right)+8$

$3 x+3=2\left(\frac{3+x}{3}\right)+8$

$3 x+3=\left(\frac{6+2 x}{3}\right)+\frac{24}{3}$

3(3x + 3) = 6 + 2x + 24

9x + 9 = 30 + 2x

7x = 21

x = 21/7

x = 3

Putting value of x in eq2

$\frac{3+3}{6}=y$

$y=1$

(v) $4^{x-1} \times(0.5)^{3-2 x}=(1 / 8)^{x}$

$2^{2 x-2} \times(5 / 10)^{3-2 x}=\left(1 / 2^{3}\right)^{x}$

$2^{2 x-2} \times(1 / 2)^{3-2 x}=2^{-3 x}$

$2^{2 x-2} \times 2^{-3+2 x}=2^{-3 x}$

2x − 2 − 3 + 2x = −3x [By equating exponents]

4x + 3x = 5

7x = 5x = 5/7

(vi) $\sqrt{\frac{\mathrm{a}}{\mathrm{b}}}=\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{1-2 \mathrm{x}}$

$(a / b)^{1 / 2}=(a / b)^{-(1-2 x)} 1 / 2=-1+2 x$ [By equating exponents]

1/2 + 1 = 2x

2x = 3/2

x = ¾