SChand CLASS 9 Chapter 6 Indices/Exponent Exercise 6(D)

 Exercise 6 D

Question 1

$2^{x+1}=4^{x-3}$

Sol:

$2^{x+1}=4^{x-3}$

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

$=2^{x+1}=2^{2 x-6}$

Comparing both sides of powers

$\begin{aligned}& x+1=2 x-6 \\\Rightarrow & 2 x-x=1+6 \\=& x=7\end{aligned}$

So= x=7 Ans 

Question 2

$x^{\frac{-3}{4}}=\frac{1}{8}$

Sol:

$\begin{aligned} x^{-3 / 4} &=\frac{1}{8} \\ &=\frac{1}{8}=\frac{1}{2^{3}}=2^{-3} \end{aligned}$

$x \frac{1}{4} \times(-3)=2^{-3}$

$=\left(x^{1 / 4}\right)^{-3}=2^{-3}$

=$\left(x^{1 / 4}\right)=2$  doing comparing.

$=x=2^{4}$

$=x=16$

So= x=16 Ans 

Question 3

$(x-1)^{\frac{2}{3}}=25$

Sol:

$\begin{aligned}(x-1)^{2 / 3} &=25 \\ &=25=5^{2} \\(x-1)^{2 / 3} &=\left(5^{3}\right)^{2 / 3} \\ x-1 &=5^{3} \text { doing comparing  } \\ x-1 &=125 \end{aligned}$

x-1=125

x=125+1

x=126 Ans.

Question 4

$2^{5 x+3}=8^{x+3}$

Sol:

$2^{5 x+3}=8^{x+3}$

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

$=2^{5 x+3}=2^{3 x+9}$

$5x+3=3x+3$ 

$5x-3x=9-3$

2x=6

x=3

Question 5

$\left(\sqrt{\frac{5}{7}}\right)^{x-1} &=\left(\frac{125}{343}\right)^{-1}$

Sol:

$\begin{aligned}\left(\sqrt{\frac{5}{7}}\right)^{x-1} &=\left(\frac{125}{343}\right)^{-1} \\ \Rightarrow\left(\frac{5}{7}\right)^{\frac{x-1}{2}} &=\left(\frac{5^{3}}{7^{3}}\right)^{-1}=\left(\frac{5}{7}\right)^{-3} \end{aligned}$

if comparing both sides power

$\begin{aligned} & \frac{x-1}{2}=-3 \\=& x-1=-6 \\ \Rightarrow & x=-6+1 \\=& x=-5 \\ & x_{0}, x=-5 \text { Ans. } \end{aligned}$

Question 6

$11^{3-4 x}=\left(\sqrt{\frac{1}{121}}\right)^{-2}

Sol:

=$11^{3-4 x}=\left(\sqrt{\frac{1}{121}}\right)^{-2}=\left(\sqrt{\frac{1}{(11)^{2}}}\right)^{-2}$

=$11^{3-4 x}=11 \frac{(-2) \times(-2)}{2}$

=$11^{3-4 x}=11^{2}$

Comparing both side of power

$\begin{aligned}3-4 x &=2 \\\Rightarrow-4 x &=2-3 \quad=-1\end{aligned}$

$\begin{aligned}&-4 x=-1 \\&x=\left(+\frac{1}{4}\right)\end{aligned}$

So $x=\frac{1}{4}$ Arus.

Question 7

Solve for x,$(\sqrt[3]{4})^{2 x+\frac{1}{2}}=\frac{1}{32}$ 

Sol :

$(\sqrt[3]{4})^{2 x+\frac{1}{2}}=\frac{1}{32}$ 

$\left(\sqrt[3]{(2)^{2}}\right)^{2 x+\frac{1}{2}}=\frac{1}{2^{5}}$ $\left\{\therefore 2^{5}=2 \times 2 \times 2 \times 2 \pi 2=32\right\}$

$\Rightarrow \quad\left[2^{3 / 3}\right]^{2 x+1 / 2}=2^{5}$

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

If comparing 

$\frac{2}{3}\left(2 x+\frac{1}{2}\right)=-5$

$\frac{4}{3} x+\frac{1}{3}=-5$

So $\begin{aligned} & 4 x+1=-15 \\ & 4 x=-13-1=-16 \\ \therefore & x=\frac{-16}{4} \\ \therefore & x=-4 \text { Ans } \end{aligned}$

Question 8

Find the value of .Y if  $\sqrt{\frac{p}{q}}=\left(\frac{q}{p}\right)^{1-2 x}$

Sol:

$\begin{aligned} & \sqrt{\frac{p}{q}}=\left(\frac{q}{p}\right)^{1-2 x} \\ &\left(\frac{p}{q}\right)^{1 / 2}=\left(\frac{p}{2}\right)^{-(1-2 x)} \\=&\left(\frac{p}{q}\right)^{2 x-1} \\ & 2 x-1=\frac{1}{2} \\ & 2 x=\frac{1}{2}+1=\frac{3}{2} \end{aligned}$

$\Rightarrow x=\frac{3}{2 \times 2}=\frac{3}{4}$

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

Question 9

Solve for x, $2^{3}\left(5^{\circ}+3^{2 x}\right) &=8 \frac{8}{27}$

Sol:

$\begin{aligned} 2^{3}\left(5^{\circ}+3^{2 x}\right) &=8 \frac{8}{27} \quad \therefore[2778+8] \\ 8\left(1+3^{2 x}\right) &=\frac{224}{27} \quad\left[5^{\circ}=1\right] \end{aligned}$

=$1+3^{2 x}=\frac{224}{27} \times \frac{1}{-8}$

$=\frac{28}{27}$

=$3^{2 x}=\frac{28}{27}-1$

=$\frac{28-27}{27}=\frac{1}{27}$

$3^{2 x}=\frac{1}{3^{2}}=3^{-3}$

$2 x=-3$

$x=-3 / 2 \quad So \quad x=\frac{-3}{2}$ Ans

Question 10

Solve for x, $\sqrt{\left(8^{0}+\frac{2}{3}\right)}=(0.6)^{2-3 x}$

Sol:

$\sqrt{\left(8^{0}+\frac{2}{3}\right)}=(0.6)^{2-3 x}$ 

$\left(8^{\circ}=1\right)$

$\sqrt{1+\frac{2}{3}}=\left(\frac{6}{10}\right)^{2-3 x}$

$\sqrt{\left(\frac{5}{3}\right)}=\left(\frac{3}{5}\right)^{2-3 x}$

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

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

$-2+3 x=\frac{1}{2} \quad$ [comparing in power]

$=3 x=\frac{1}{2}+2=\frac{5}{2}$

$=x=\frac{5}{2 \times 3}=\frac{5}{6}$

$\therefore x=\frac{5}{6}$

Question 11

$3^{2 x+4}+1=2 \cdot 3^{x+2}$

Sol:

$3^{2 x+4}+1=2 \cdot 3^{x+2}$

$3^{2 x} \cdot 3^{4}+1=2 \cdot 3^{x} \cdot 3^{2}$

$81 \cdot 3^{2 x}+1=18 \cdot 3^{x}$

$81 \cdot 32 x-18 \cdot 3^{x}+1=0$

$3 x=a$,

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

$\begin{aligned} \therefore & 81 a^{2}-18 a+1=0 \\ &(9 a)^{2}-2 \times 9 a+1=0 \\ &(9 a-1)^{2}=0 \\ &: 9 a-1=0 \\ \Rightarrow & 9 a=1=0 \\ \Rightarrow & 9 a=1 \\ \Rightarrow & 9 \cdot 3^{x}=1 \end{aligned}$

$3^{x}=\frac{1}{9}=3^{-2}$  If comparing both sides power

x=-2 ans.

Question 12

$5^{2 x+1}=6 \cdot 5^{x}-1$

Sol:

$\begin{aligned} & 5^{2 x+1}=6 \cdot 5^{x}-1 \\=& 5^{2 x} \cdot 5^{2}-6 \cdot 5^{x}+1=0 \\ & 5 \cdot 5^{2 x}-6 \cdot 5^{x}+1=0 \end{aligned}$

$\begin{aligned} 5^{x} &=a \\ \text { and } 5^{2 x} &=a^{2} \end{aligned}$

$\begin{aligned} \therefore \quad 5 a^{2}-6 a+1 &=0 \\ 5 a^{2}-5 a-a+1 &=0 \\ 5 a(a-1)-1(a-1) &=0 \\(a-1)(5 a-1) &=0 \end{aligned}$

$\Rightarrow$

=$\begin{aligned}a-1 &=0 \\a &=2\end{aligned}$

$5 a-1=0$

$5 a=1$

$a=\frac{1}{5}$

case (i)

$\begin{aligned}&a=1,5^{x}=1=5^{\circ} \\&\therefore x=0\end{aligned}$

case (ii) $a=\frac{1}{3}, \quad 5^{x}=\frac{1}{5}=5^{-L}$ $\therefore x=-1$

$\therefore x=-1$

$\begin{aligned} \text { Hence } x &=0 \\ x &=-1 \end{aligned}$

Question 13

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

Sol:

$\begin{aligned} & 2^{2 x}-2^{x+3}=-2^{4} \\ & 2^{2 x}-2^{x} \cdot 2^{3}+2^{4}=0 \\ & 2^{2 x}-8 \cdot 2 x+16=0 \\ \because & 2^{x}=a, \quad 2^{2 x}=a^{2} \\ & a^{2}-8 a+16=0 \\ \Rightarrow &(a)^{2}-2 \times a \times 4+(4)^{2}=0 \end{aligned}$

$=(a-4)^{2}=0$

$a-4=0$

$a=2 x=4$

$2 x=2^{2}$(Comparing)

x=2 Ans

Question 14

$9^{x}=3^{y-2}, 81^{y}=3^{2} \times(27)^{x}$

Sol:

$\begin{aligned}&9^{x}=3^{y-2}, 81^{y}=3^{2} \times(27)^{x} \\&{\left[(3)^{2}\right]^{x}=(3)^{y-2}} \\&\Rightarrow 3^{2 x}=3 y^{-2}\end{aligned}$

comparing both sides power.

$2 x=y-2$

$y=2 x+2$............(equal -1)

$\therefore 81^{y}=3^{2 \times(27)^{x}}$

$\Rightarrow\left(3^{4}\right)^{y}=3^{2}\left(3^{3}\right)^{x} \quad(3 \times 3 \times 3 \times 3=81)$

$\begin{aligned}\left(3^{4}\right)^{y} &=3^{2} \times 3^{3 x} \\-3^{4 y} &=3^{3 x+2} \\ \therefore 4 y &=3 x+2 \end{aligned}$........(equal -2)

from equation -1

$a(2 x+2)=3 x-2$

$8 x+8=3 x+2$

$3 x-3 x=2-8$

$5 x=-6$

$x=\frac{-6}{5}$

$\therefore y=2 x+2=2 \times\left(-\frac{6}{5}\right)+2$

$=\frac{-12}{5}+2=\frac{-12+10}{5}=\frac{-2}{5}$

$\therefore x=\frac{-6}{5}$

$\therefore y=\frac{-2}{5}$

Hence proved

Question 15

$2^{1-\frac{x}{2}}=4^{y}$ , $7^{1+x} \times(49)^{-2 y}=1$

Sol:

$2^{1-\frac{x}{2}}=4^{y} \Rightarrow 2^{1-\frac{x}{2}}=2^{2 y}$

$1-\frac{x}{2}=2 y$

$2-x=4 y$

$4 y+x=2$

$x=2-4 y$...........(eq 2)

as $7^{1+x} \times(49)^{-2 y}=1$

$7^{1+x} \cdot\left((7)^{2}\right)^{-2 y}=1$.........(1=7)

$\begin{aligned} 7^{1+x} \cdot 7^{-4 y} &=7^{\circ} \\ 7^{1+x-4 y} &=7^{\circ} \\ \therefore \quad 1+x-4 y &=0 \\ 1+(2-4 y) &-4 y=0 \end{aligned}$

$\therefore \quad 1+x-4 y=0$

$\Rightarrow 1+(2-4 y)-4 y=0$

From equation 1

$x=2-4 y$

$1+2-4 y-4 y=0$

$3-8 y=0$

$8 y=3 \Rightarrow y=3 / 8 .$

$x=2-4 y=2-4 \times \frac{3}{8}$

$=2-3 / 2$

$=\frac{4-3}{2}=\frac{1}{2}$

Hence 

$x=\frac{1}{2}$

$y=3 / 8$

Question 16

$2^{x}=16 \times 2^{y},(27)^{x}=9 \times 3^{2 y}$

Sol:

$\begin{aligned} 2^{x} &=16 \times 2^{y},(27)^{x}=9 \times 3^{2 y} \\ 2^{x} &=16 \times 2^{y} \\ 2^{x} &=2^{4} \times 2 y \\ 2^{x} &=2^{4}+y \end{aligned}$

Comparing both sides power

$\therefore x=4+y$............(eq-1)

 $\begin{aligned}(27)^{x} &=9 \times 3^{2 y} \\\left(3^{3}\right)^{x} &=3^{2} \times 3^{2} y \\ \therefore 3^{3 x} &=3^{2 y+2} \end{aligned}$

$3 x=2 y+1$ (3 Common )

$3(4+y)=2 y+2$ From Eq-1

$12+3 y=2 y+2$

$3 y-2 y=2-12$

$y=-10$

$\begin{aligned} \therefore x=4+y &=4-10 \\ &=-6 \end{aligned}$

$\left[\begin{array}{l}x=-6 \\ y=-10\end{array}\right]$ Hence proved