ML Aggarwal Solution Class 9 Chapter 9 Logarithms Exercise 9.1

 Exercise 9.1

Question 1

Convert the following to logarithmic form:

(i) 52 = 25

⇒log525=2


(ii) a5 =64

⇒loga64=5


(iii) 7x =100

⇒log7100=x


(iv) 9° = 1

⇒log91=0


(v) 61= 6

⇒log66=1


(vi) 3-2 =$\frac{1}{9}$

⇒$\log _{3} \frac{1}{9}=-2$


(vii) 10-2 = 0.01

⇒log100.01=-2


(viii) $(81)^{\frac{3}{4}} = 27$

⇒log8127$=\frac{3}{4}$


Question 2

Convert the following into exponential form:

(i) log2 32 = 5

⇒25=32


(ii) log3 81=4

⇒34=81


(iii) log3 $\frac{1}{3}$= -1

⇒$3^{-1}=\frac{1}{3}$


(iv) log3 4= $\frac{2}{3}$

⇒$(8)^{\frac{2}{3}}=4$


(v) log8 32= $\frac{5}{3}$

⇒$(8)^{\frac{5}{3}}=32$


(vi) log10 (0.001) = -3

⇒10-3=0.001


(vii) log2 0.25 = -2

⇒2-2=0.25


(viii) loga ($\frac{1}{a}$) =-1

⇒$a^{-1}=\frac{1}{a}$


Question 3

By converting to exponential form, find the values of:

(i) log2 16

⇒Let, log2 16=x

⇒(2)x=16

⇒(2)x=2×2×2×2

⇒(2)x=(2)4

∴x=4


(ii) log5 125

⇒Let, log5 125=x

⇒(5)x=125

⇒(5)x=5×5×5

⇒(5)x=(5)3

∴x=3


(iii) log4 8

⇒Let, log4 8=x

⇒(4)x=8

⇒(2×2)x=2×2×2

⇒(2)2x=(2)3

⇒2x=3

∴$x=\frac{3}{2}$


(iv) log9 27

⇒ log9 27=x

⇒9x=27

⇒(3×3)x=3×3×3

⇒(3)2x=33

⇒2x=3

∴$x=\frac{3}{2}$


(v) log10 (.01)

⇒log10 (.01)=x

⇒(10)x=0.01

⇒$(10)^{x}=\frac{1}{100}$

⇒$(10)^{x}=\frac{1}{10} \times \frac{1}{10}$

⇒$(10)^{x}=\frac{1}{(10)^{2}}$

⇒(10)x=(10)-2

∴x=-2


(vi) log7 $\frac{1}{7}$

⇒ log7 $\frac{1}{7}=x$

⇒$(7)^{x}=\frac{1}{7}$

⇒(7)x=(7)-1

∴x=-1


(vii) log5 256

⇒log5 256=x

⇒(0.5)x=256

⇒$\left(\frac{5}{10}\right)^{x}=256$

⇒$\left(\frac{1}{2}\right)^{x}=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

⇒(2)-x=(2)8

⇒-x=8

∴x=-8


(viii) log2 0.25

⇒Let, log2 0.25=x
⇒(2)x=0.25
⇒(2)x=$\frac{25}{100}$
⇒$(2)^{x}=\frac{1}{4}$
⇒(2)x=(2)-2
∴x=-2


Question 4

Solve the following equations for x.

(i)  log3 x=2
⇒32
⇒9=x
∴x=9

(ii) logx 25=2
⇒x2=25
⇒x2=5×5 
⇒x2=52
∴x=5

(iii) log10 x=-2
⇒(10)-2=x
⇒x=(10)-2
⇒$x=\frac{1}{(10)^2}$
⇒$x=\frac{1}{100}$
∴x=0.01

(iv) log4 x=$\frac{1}{2}$
⇒$(4)^{\frac{1}{2}}=x$
⇒$x=(4)^{\frac{1}{2}}$
⇒$x=(2\times 2)^{\frac{1}{2}}$
⇒$x=(2)^{2\times \frac{1}{2}}$
⇒x=(2)1
∴x=2

(v) logx 11=1
⇒(x)1=11
∴x=2

(vi) $\log _{x} \frac{1}{4}=-1$
⇒$(x)^1=\frac{1}{4}$
⇒(x)1=(4)-1
∴x=4

(vii) $\log _{81} x=\frac{3}{2}$
∴$x=811^{\frac{3}{2}}$
$=(3^4)^{\frac{3}{2}}=3^{4\times \frac{3}{2}}$
⇒36
=3×3×3×3×3×3=729

(viii) log9 x=2.5
⇒$\log _{9} x=2.5=\frac{5}{2}$
∴$x=(9)^{\frac{5}{2}}=\left(3^{2}\right)^{\frac{5}{2}}=3^{2 \times \frac{5}{2}}=3^{5}$
=3×3×3×3×3=243

(ix) log4 x=-1.5
⇒$-1.5=\frac{-3}{2}$
∴$x=(4)^{\frac{-3}{2}}=\left(2^{2}\right)^{-3}=2^{2 \times\left(\frac{-3}{2}\right)}$
$=2^{-3}=\frac{1}{2^{3}}=\frac{1}{2 \times 2 \times 2}=\frac{1}{8}$

(x) $\log _{\sqrt{5}} x=2$
⇒$(\sqrt{5})^{2}=x$
⇒$(5)^{2 \times \frac{1}{2}}=x$
⇒(5)1=x
∴x=5

(xi) logx 0.001=-3
⇒$(x)^{-3}=\frac{1}{1000}$
⇒$(x)^{-3}=\frac{1}{(10)^{3}}$
∴(x)-3=10-3
∴x=10

(xii) $\log _{\sqrt{3}} (x+1)=2$
⇒$(\sqrt{3})^{2}=x+1$
⇒3=x+1
⇒x+1=3
⇒x=3-1
∴x=2

(xiii) $\log _{4}(2 x+3)=\frac{3}{2}$
⇒$(4)^{\frac{3}{2}}=2 x+3$
⇒$(2 \times 2)^{\frac{3}{2}}=2 x+3$
⇒$(2)^{2 \times  \frac{3}{2}}=2 x+3$
⇒$(2 \times 2)^{\frac{3}{2}}=2 x+3$
⇒2×2×2=2x+3
⇒8=2x+3
⇒2x=8-3
∴2x=5
∴$x=\frac{5}{2}$

(xiv) $\log _{\sqrt[3]{2}} x=3$
⇒$(\sqrt[3]{2})^{3}=x$
⇒$\left[(2)^{\frac{1}{3}}\right]^{3}=x$
⇒$(2)^{\frac{1}{3}} \times 3=x$
⇒(2)1=x
∴x=2

(xv) $\log _{2}\left(x^{2}-1\right)=3$
⇒(2)3=x2-1
⇒2×2×2=x2-1
⇒8=x2-1
⇒x2-1=8
⇒x2=8+1
⇒x2=9
∴x=±3

(xvi) log x=-1
⇒(10)-1=x
⇒x=(10)-1
∴$x=\frac{1}{10}$

(xvii) log (2x-3)=1
⇒(10)-1=2x-3
⇒10=2x-3
⇒2x=10+3
⇒2x=13
∴$x=\frac{13}{2}=6 \frac{1}{2}$

(xviii) $\log x=-2,0, \frac{1}{3}$
⇒log x=-2
⇒(10)-2=x
⇒$\frac{1}{100}=x$
⇒$x=\frac{1}{100}$

When log x=0
⇒(10)0=x
⇒x=1

When $\log x=\frac{1}{3}$
⇒$(10)^{\frac{1}{3}}=x$
⇒$x=\sqrt[3]{10}$

Hence, $x=\frac{1}{100}, 1 \sqrt[3]{10}$


Question 5

Given  log10 a=b, express 102b-3 in terms of a.

Sol :

Given  log10 a=b

⇒(10)h=a

Now $10^{2h-3}=\frac{(10)^{2 b}}{(10)^{3}}=\frac{\left(10^{b}\right)^{2}}{10 \times 10 \times 10}=\frac{\left(10^{b}\right)^{2}}{1000}$

$=\frac{a^{2}}{1000}$


Question 6

Given log10 x= a, log10 y = b and log10 z =c,

(i) Write down 102a-3 in terms of x.

(ii) Write down 103b-1 in terms of y.

(iii) If log10 P $=2 a+\frac{b}{2}-3 c$, express P in terms of x, y and z.

Sol :

Given that log10 x=a
⇒(10)a=x..(1)
⇒log10 y=b⇒(10)b=y...(2)
⇒log10 z=c⇒(10)c=z...(3)

(i) 102a-3=$\frac{(10)^{2 a}}{(10)^{3}}=\frac{\left(10^{a}\right)^{2}}{10 \times 10 \times 10}=\frac{(x)^{2}}{1000}=\frac{x^{2}}{1000}$

(ii) $10^{3b-1}=\frac{(10)^{3 b}}{(10)^{1}}=\frac{\left(10^{b}\right)^{3}}{10}=\frac{(y)^{3}}{10}=\frac{y^{3}}{10}$

(iii) log10 P=$2 a+\frac{b}{2}-3 c$

Substitute the value of a,b and c from equation (1), (2) and (3)
⇒log10 P$=2 \log _{10} x+\frac{1}{2} \log _{10} y-3 \log _{10} z$
⇒$\log _{10} \mathrm{P}=\log _{10}(x)^{2}+\log _{10}(y)^{\frac{1}{2}}-\log _{10}(2)^{3}$
⇒$\log _{10} P=\log _{10}\left(x^{2} \times y^{\frac{1}{2}}\right)-\log _{10} z^{3}$
⇒$\log _{10} P=\log _{10}\left(\frac{x^{2} \sqrt{y}}{z^{3}}\right)$
⇒$P=\frac{x^{2} \sqrt{y}}{z^{3}}$

Question 7

If log10 = a and log10 y = b, find the value of xy.

Sol :

Given : That log10 x=a
⇒(10)a=x and (10)b=y

Then xy=(10)a×(10)b=(10)a+b


Question 8

Given log10 a = m and log10 b = n, express $\frac{a^{3}}{b^{2}}$ in terms of m and n.

Sol :

Given log10 a = m and log10 b = n,

Then (10)m= a and (10)n =b

$\frac{a^{3}}{b^{2}}=\frac{\left(10^{m}\right)^{3}}{\left(10^{n}\right)^{2}}=\frac{(10)^{3 m}}{(10)^{2 n}}$

=(10)3m-2n


Question 9

Given log10 a = 2a and log10 y $=\frac{-b}{2}$

(i) Write 10a in terms of x.

(ii) Write 102b+1 in terms of y.

(iii) If log10 P= 3a -2b, express P in terms of x and y .

Sol :
Given that log10 a = 2a
⇒(10)2a=x

and log10 y$=\frac{b}{2}$ ,

⇒$(10)^{\frac{b}{2}}=y$

(i) $10^{a}=\left(10^{2a}\right)^{\frac{1}{2}}=(x)^{\frac{1}{2}}=\sqrt{x}$


(ii) (10)2b+1=(10)2b×(10)1

$=10^{4}\left(\frac{b}{2}\right) \times 10^{1}$

$=\left(10^{\frac{b}{2}}\right)^{4} \times 10=y^{4} \times 10=10 y^{4}$


(iii) log10 P=3a-2b

⇒log10 P$=\frac{3}{2}(2 a)-4\left(\frac{b}{2}\right)$

⇒log10 P$=\frac{3}{2}\left(\log _{10} x\right)-4=\left(\log _{10} y\right)$

⇒log10 P$=\log _{10}(x)^{\frac{3}{2}}-\log _{10} y^{4}$

⇒$\log _{10} P=\log _{10}\left(\frac{(x)^{\frac{3}2{}}}{y^{4}}\right)$

⇒$P=\frac{(x)^{\frac{3}{2}}}{y^{4}}$


Question 10

If log2 y = x and log3 z = x, find 72x in terms of y and z.

Sol :
⇒log2 y = x and log3 z = x

⇒y=2x and z=3x..(i)

⇒72x=(2×2×2×3×3)x

=(23×32)x

=(2x)3×(3x)2

=y3.z2  [From (i)]

Hence , 72x=y3.z2 


Question 11

If log2 x = a and log5 y = a, write 1002a-1 in terms of x and y.

Sol :
⇒log2 x =a and log5 y =a

∴x=2a and y=5a

1002a-1=(2×2×5×5)2a-1

=(22×52)2a-1=24a-2×54a-2

$=\frac{2^{4 a}}{2^{2}} \times \frac{5^{4 a}}{5^{2}}$

$=\frac{\left(2^{a}\right)^{4} \times\left(5^{a}\right)^{4}}{4 \times 25}$

$=\frac{x^{4} \times y^{4}}{100}$

$=\frac{x^{4} y^{4}}{100}$

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference be...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more