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

⇒log_a64=5

(iii) 7^x =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) log_a ($\frac{1}{a}$) =-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

⇒9^x=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) log_x 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) log_x 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) log_x 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 10^2b-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 10^2a-3 in terms of x.

(ii) Write down 10^3b-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) 10^2a-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 10^a in terms of x.

(ii) Write 10^2b+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 72^x in terms of y and z.

Sol :

⇒log₂ y = x and log₃ z = x

⇒y=2^x and z=3^x..(i)

⇒72^x=(2×2×2×3×3)^x

=(2³×3²)^x

=(2^x)³×(3^x)²

=y³.z²  [From (i)]

Hence , 72^x=y³.z² 

Question 11

If log₂ x = a and log₅ y = a, write 100^2a-1 in terms of x and y.

Sol :

⇒log₂ x =a and log₅ y =a

∴x=2^a and y=5^a

100^2a-1=(2×2×5×5)^2a-1

=(2²×5²)^2a-1=2^4a-2×5^4a-2

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

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

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

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