ML Aggarwal Solution Class 9 Chapter 9 Logarithms Exercise 9.2

 Exercise 9.2

Question 1

Simplify the following :

(i) log a3-log a2
Sol :
⇒log a3-log a2
$=\log \left(\frac{a^{3}}{a^{2}}\right)$ (Quotient Law)
=log a

(ii) log a3÷log a2
Sol :
=3log a÷2log a (Power Law)
$=\frac{3 \log a}{2 \log a}=\frac{3}{2}$

(iii) $\frac{\log 4}{\log 2}$
Sol :
$=\frac{\log (2 \times 2)}{\log 2}=\frac{\log 2}{\log 2}$
$=\frac{2 \log 2}{\log 2}$ (Power Law)
=2(1)=2

(iv)$\frac{\log 8 \log 9}{\log 27}$
Sol :
$=\frac{\log 2^{3} \cdot \log 3^{2}}{\log 3^{3}}$
$=\frac{(3 \log 2) \cdot(2 \log 3)}{(3 \log 3)}$ (Power Law)
$=\frac{(\log 2) \cdot(2)}{1}$
=2 log 2=log 22
=log 4

(v) $\frac{\log 27}{\log \sqrt{3}}$
Sol :
$=\frac{\log (3 \times 3 \times 3)}{\log (3)^{\frac{1}{2}}}$
$=\frac{\log (3)^{3}}{\log (3)^{\frac{1}{2}}}$
$=\frac{3 \log 3}{\frac{1}{2} \log 3}$ (Power Law)
$=\frac{3 \times 2}{1}\left(\frac{\log 3}{\log 3}\right)$
=6(1)=6


(vi) $\frac{\log 9-\log 3}{\log 27}$
Sol :
$=\frac{\log (3 \times 3)-\log 3}{\log (3 \times 3 \times 3)}$

$=\frac{\log 3^{2}-\log 3}{\log 3^{3}}$

$=\frac{2 \log 3-\log 3}{3 \log 3}$ (Power Law)

$=\frac{\log 3}{3 \log 3}=\frac{1}{3}$


Question 2

Evaluate the following:

(i) $\log (10 \div \sqrt[3]{10})$

Sol :

$=\log \left((10)^{1} \div(10)^{\frac{1}{3}}\right)$

$=\log \left(10^{1-\frac{1}{3}}\right)$

$=\log \left(10^{\frac{2}{3}}\right)$

$=\frac{2}{3} \log 10=\frac{2}{3}(1)=\frac{2}{3}$


(ii) $2+\frac{1}{2} \log \left(10^{-3}\right)$

Sol :

$=2+\frac{1}{2} \times(-3) \log 10$

$=2-\frac{3}{2} \log 10=2-\frac{3}{2}(1)$

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


(iii) $\log 5+\log 8-\frac{1}{2} \log 4$

Sol :

$=\log (5)^{2}+\log 8-\frac{1}{2} \log (2)^{2}$

$=\log 25+\log 8-\frac{1}{2} \times 2 \log 2$

$=\log 25+\log 8-\log 2=\log \left(\frac{25 \times 8}{2}\right)$

$=\log \left(\frac{25 \times 4}{1}\right)$

=log (100)

=log (10)2

=2 log 10=2(1)=2


(iv) $2 \log 10^{3}+3 \log 10^{-2}-\frac{1}{3} \log 5^{-3}+\frac{1}{2} \log 4$

Sol :

$=2 \times 3 \log 10+3(-2) \log 10-\frac{1}{3}(-3) \log 5+\frac{1}{2}$

=log (2)2

=6 log 10-6 log 10+$\frac{3}{3} \log 5+\frac{1}{2} \times 2 \log 2$

=0+1 log 5 +log 2=log 5+log 2

=log (5×2)

=log(10)=1


(v) $2 \log 2+\log 5-\frac{1}{2} \log 36-\log \frac{1}{30}$

Sol :

$=\log (2)^{2}+\log 5-\frac{1}{2} \log (6)^{2}-\log \left(\frac{1}{30}\right)$

$=\log 4+\log 5-\frac{1}{2} \times 2 \log 6-\log \frac{1}{30}$

$=\log 4+\log 5-\frac{1}{2} \times 2 \log 6-\log \frac{1}{30}$

=log 4+log 5-log 6-(log 1-log 30)

=log 4+log 5-log 6-log 1+log 30

=(log 4+log 5+log 30)-(log 6+log 1) 

=log (4×5×30)-log (6×1)

$=\log \frac{4 \times 5 \times 30}{6 \times 1}=\log \frac{4 \times 5 \times 5}{1 \times 1}$

=log 100

=log (10)2

=2 log 10

=2(1)=2


(vi) $2 \log 5+\log 3+3 \log 2-\frac{1}{2} \log 36-2 \log 10$

Sol :

=log (5)2+log 3+log (2)3$-\frac{1}{2} \log (6)^{2}-2 \log 10$
=log 25+log 3+log 8$-\frac{1}{2} \times 2 \log 6-2 \log 10$

=log 25+log 3+log 8-log 6-log (10)2

=log (25×3×8)-log 6-log 100

$=\log \left(\frac{25 \times 3 \times 8}{6 \times 100}\right)=\log \left(\frac{1 \times 3 \times 8}{6 \times 4}\right)$

$=\log \left(\frac{24}{24}\right)$

=log 1=0


(vii) $\log 2+16 \log \frac{16}{15}+12 \log \frac{25}{24}+7 \log \frac{81}{80}$

Sol :

=log 2+16(log 16-log 15)+12(log 25-log 24)+7(log 81-log 80)

=log 2+16[log (2)4-log (3×5)]+12[log (5)2-log(3×2×2×2)+7[log(3×3×3×3)-log (2)4×5]

=log 2+16[4 log 2-(log 3+log 5)]+12[2 log 5-log(3×23)]+7[log(3)4-(log 4+log 5)]

=log 2+16[4 log 4- log 3 - log 5]+12[2 log 5-log (3×23)]+7[log 34-(log 4+ log 5)]

=log 2+16[4 log 4-log 3-log 5]+12[2 logs 5 -(log 3-log 23)]+7[4 log 3-log 4-log 5]

=log 2+ 64 log 2-16 log 3-16 log 5+24 log 5-12 log 3-12 log 23+28 log 3-7 log 24-7 log 5

=log 2+ 64 log 2-16 log 3-16 log 5+24 log 5-12 log 3-36 log 2+28 log 3-28 log 2-7 log 5

=(log 2+ 64 log 2-36 log 2-28 log 2)+(-16 log 3-12 log 3+28 log 3)+(-16 log 5+24 log 5-7 log 5)+28 log 3)+(-16 log 5+24 log 5-7 log 5)

=(65 log 2-64 log 2)+(-28 log 3+28 log 3)+(-23 log 5+24 log 5)
=log 2+0+log 5=log 2+log 5
=log(2×5)
=log 10 =1

(viii) $2 \log _{10} 5+\log _{10} 8-\frac{1}{2} \log _{10} 4$

Sol :

$=\log _{10}(5)^{2}+\log _{10} 8-\log _{10}(4)^{\frac{1}{2}}$

$=\log _{10} 25+\log _{10} 8-\log _{10}(2)^{2 \times \frac{1}{2}}$

$=\log _{10} 25+\log _{10} 8-\log _{10} 2$

$=\log _{10}\left(\frac{25 \times 8}{2}\right)=\log _{10}(25 \times 4)$

$=\log _{10} 100=\log _{10}(10)^{2}$

$=2 \log _{10} 10=2(1)=2$


Question 3

Express each of the following as a single logarithm:
(i) $2 \log 3-\frac{1}{2} \log 16+\log 12$
$=2 \log 3-\frac{1}{2} \log (4)^{2}+\log 12$
$=2 \log 3-\frac{1}{2} \times 2 \log 4+\log 12$
$=2 \log 3-\log 4+\log 12$
$=\log (3)^{2}-\log 4+\log 12$
=log 9-log 4+log 12
$=\log \frac{9 \times 12}{4}=\log \left(\frac{9 \times 3}{1}\right)$
=log 27


(ii) $2 \log _{10} 5-\log _{10} 2+3 \log _{10} 4+1$
$=\log _{10}(5)^{2}-\log _{10} 2+\log _{10}(4)^{3}+\log _{10} 10$ $\left(\therefore \log _{10} 10=1\right)$
$=\log _{10} 25-\log _{10} 2+\log _{10} 64+\log _{10} 10$
$=\log _{10}(25 \times 64 \times 10)-\log _{10} 2$
$=\log _{10}(16000)-\log _{10} 2$
$=\log _{10}\left(\frac{16000}{2}\right)=\log _{10} 8000$


(iii) $\frac{1}{2} \log 36+2 \log 8-\log 1.5$
$=\log (36)^{\frac{1}{2}}+\log (8)^{2}-\log 1.5$
$=\log (6)^{2 \times \frac{1}{2}}+\log 64-\log \left(\frac{15}{10}\right)$
=log 6+log 64-(log 15-log 10)
=log 6×64-log 15+log 10
=log (6×64×10)-log 15
$=\log \left(\frac{60 \times 64}{15}\right)$
=log (4×64)
=log 256


(iv) $\frac{1}{2} \log 25-2 \log 3+1$
$=\log (25)^{\frac{1}{2}}-\log (3)^{2}+\log 10$ (∵ log 10=1)
$=\log (5)^{2 \times \frac{1}{2}}-\log 9+\log 10$
=log 5-log 9+log 10
=log (5×10)-log 9
$=\log \frac{5 \times 10}{9}=\log \frac{50}{9}$


(v) $\frac{1}{2} \log 9+2 \log 3-\log 6+\log 2-2$

$=\frac{1}{2} \log 9+2 \log 3-\log 6+\log 2-2$

$=\log 9^{\frac{1}{2}}+\log 3^{2}-\log 6+\log 2-\log 100$

=log 3+log 9-log 6+log 2-log 100

$=\log \frac{3 \times 9 \times 2}{6 \times 100}$

$=\log \frac{9}{100}$


Question 4

Prove the following :

(i) log10 4 ÷ log10 2 =  log3 9
L.H.S=log10 4 ÷ log10 2 
$=\frac{2 \log _{10} 2}{\log _{10} 2}$
=2(1)=2

R.H.S=log3 9
=log3 (3)2 
=2(1)=2

Hence, proved L.H.S=R.H.S

(ii) log10 25 + log10 4 = log5 25
L.H.S=log10 25 + log10 4 
=log10 25×4 
=log10 100
=log10 102

=2 log10 10

=2×1=2 (∵log10 a=1)

Hence, L.H.S=R.H.S


Question 5

If x = 100)a , y = (10000)b and z = (10)c, express $\log \frac{10 \sqrt{y}}{x^{2} z^{3}}$ in terms of a,b,c

Sol :
Given that x=(100)a=[(10)2]a

=(10)2a


y=(10000)b=[(10)4]b

=(10)4b


z=(10)c=(10)c

Now, $\log \frac{10 \sqrt{y}}{x^{2} z^{3}}$

=(log 10+log √y)-(log x2+logz3)

$=\left(1+\log (y)^{\frac{1}{2}}\right)-\left(\log (x)^{2}+\log (z)^{3}\right)$ [∵ log 10=1]

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

Substituting the value of x, y and z, we get

$=\left(1+\frac{1}{2} \log (10)^{4 b}\right)-\left(2 \log (10)^{2 a}+3 \log (10)^{c}\right)$

$=\left(1+\frac{1}{2} \times 4 \mathrm{~b} \log 10\right)-(2 \times 2 a \log 10+3 \times c \log 10)$

$=\left(1+\frac{1}{2} \times 4 \mathrm{~b} \times 1\right)-(2 \times 2 a \times 1+3 \times c \times 1)$ [∵log 10=1]

=(1+2b)-(4a+3c)

=1+2b-4a-3c

=1-4a+2b-3c


Question 6

If a = log10 x, find the following in terms of a :

(i) x
(ii) $\log _{10} \sqrt[5]{x^{2}}$
(iii) $\log _{10} 5 x$

Sol :

(i) Given that,

⇒a=log10 x

⇒(10)a=x

⇒x=(10)a


(ii) $\log _{10} \sqrt[5]{x^{2}}$

$=\log _{10}\left(x^{2}\right)^{\frac{1}{5}}$

$=\log _{10}(x)^{\frac{2}{5}}$

$=\frac{2}{5} \log _{10} x$

$=\frac{2}{5}(a)=\frac{2}{5} a$


(iii) x=(10)a=log10 5x

=log10 5(10)a

=log10 5+log10 10

=log10 5+a(1)

=a+log10 5


Question 7

If $a=\log \frac{2}{3}, b=\log \frac{3}{5}$ and $c=2 \log \sqrt{\frac{5}{2}}$

Find the value of

(i) a+b+c

(ii) 5a+b+c

Sol :

Given that :

$a=\log \frac{2}{3}, b=\log \frac{3}{5}, c=2 \log \sqrt{\frac{5}{2}}$

(i) a+b+c$=\log \frac{2}{3}+\log \frac{3}{5}+2 \log \sqrt{\frac{5}{2}}$

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

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

=log 2-log 3+log 3-log 5+log \frac{5}{2}$

=log 2+(-log 3+log 3)-log 5+log 5-log 2$

=(log 2-log 2)+0+(log 5-log 5)

=0+0+0=0


(ii) 5a+b+c=50

=1


Question 8

If $x=\log \frac{3}{5}, y=\log \frac{5}{4}$ and $z=2 \log \frac{\sqrt{3}}{2}$ Find the value of 

(i) x+y-z 

(ii) 3x+y-z

Sol :

$x=\log \frac{3}{5}, y=\log \frac{5}{4}, z=2 \log \frac{\sqrt{3}}{2}$

∴x=log 3-log 5

y=log 5-log 4

$z=\log \left(\frac{\sqrt{3}}{2}\right)^{2}=\log \frac{3}{4}$

=log 3-log 4

(i) Now, x+y-z=log 3-log 5+log 5-log -log 3-log 4

=0

(ii) 3x+y-3

=30=1


Question 9

If x = log1012, y = log4 2×log10 9 and z = log10 0.4, find the values of

(i) x-y-z

(ii) 7x-y-z

Sol :
x = log10 12, y = log4 2×log10 9 and z = log10 0.4

(i) x-y-z

= log10 12-log4 2×log10 9- log10 0.4

= log10 (3×4)-log4 41/2×log10 32- log10 $\frac{4}{10}$

=$=\log _{10} 3+\log _{10} 4-\frac{1}{2} \log _{4} 4 \times 2 \log _{10} 3-\left(\log _{10} 4-\log _{10} 10\right)$

$=\log _{10} 3+\log _{10} 4-\frac{1}{2} \times 1 \times 2 \log _{10} 3-\log _{10} 4+1$

$=\log _{10} 3+\log _{10} 4-\log _{10} 3-\log _{10} 4+1$

=1


(ii) 7x-y-z=71=0


Question 10

If log V + log 3 = log π + log 4 + 3 log r, find V in terns of other quantities.

Sol :
⇒$\log V+\log 3=\log \pi+\log 4+\log (r)^{3}$
⇒$\log (\mathrm{V} \times 3)=\log \left(\pi \times 4 \times r^{3}\right)$

⇒$\log (3 \mathrm{~V})=\log 4 \pi r^{3}$

⇒$3 \mathrm{~V}=4 \pi r^{3}$

⇒$V=\frac{4}{3} \pi r^{3}$


Question 11

Given 3 (log 5 – log 3) – (log 5-2 log 6) = 2 – log n , find n.

Sol :
Given that 3(log 5-log 3)-(log 5-2 log 6)= 2-log n

⇒3 log 5-3  log 3- log 5+2 log 6=2- log n 

⇒2 log 5-3 log 3+2 log 6=2-log n

⇒$\log (5)^{2}-\log (3)^{3}+\log (6)^{2}=2(1)-\log n$

⇒log 25-log 27+log 36=2 log 10- log n  [∵log 10=1]

⇒log n=2 log 10-log 25+log 27-log 36

⇒log n=log (10)2-log 25+log 27-log 36

⇒log n=log 100-log 25+log 27-log 36

⇒log n=(log 100+log 27)-(log 25+log 36)

⇒log n=log (100×27)-log (25×36)

⇒$\log n=\log \left(\frac{100 \times 27}{25 \times 36}\right)$

⇒$\log n=\log \left(\frac{4 \times 27}{1 \times 36}\right)$

⇒$\log n=\log \left(\frac{1 \times 27}{1 \times 9}\right)$

⇒log n=log 3

⇒n=3


Question 12

Given that log10 y + 2 log10 x= 2, express y in terms of x.

Sol :
⇒log10 y + 2 log10 x= 2
⇒log10 y + log10 x2= 2
⇒log10 yx2= 2
⇒log10 yx2= 2 log10 10
⇒log10 (yx2)= 2 log10 (10)2
⇒yx2=(10)2
⇒yx2=100
⇒$y=\frac{100}{x^{2}}$


Question 13

Express log10 2+1 in the from log10 x.

Sol :
⇒log10 2+1=log10 2+log10 10 (∵ log10 10=1)
⇒log10 2×10=log10 20

Question 14

If $a^{2}=\log _{10} x, b^{3}=\log _{10} y$ and $\frac{a^{2}}{2}-\frac{b^{2}}{3}=\log _{10} z$ express z in terms of x and y
Sol :
Given that
⇒$a^{2}=\log _{10} x, b^{3}=\log _{10} y$
⇒We have $\frac{a^{2}}{2}-\frac{b^{2}}{3}=\log _{10} z$
⇒$\frac{1}{2}\left(\log _{10} x\right)-\frac{1}{3}\left(\log _{10} y\right)=\log _{10^{2}}$
⇒$\log _{10}(x) \frac{1}{2}-\log _{10}(y) \frac{1}{3}=\log _{10^{2}}$
⇒$\log _{10} \sqrt{x}-\log _{10} \sqrt[3]{y}=\log _{10} z$
⇒$\log _{10} \frac{\sqrt{x}}{\sqrt[3]{y}}=\log _{10} z$
⇒$\frac{\sqrt{x}}{\sqrt[3]{y}}=z$
Hence , $z=\frac{\sqrt{x}}{\sqrt[3]{y}}$

Question 15

Given that log m = x + y and log n = x-y, express the value of log m2n in terms of x and y.
Sol :
Given that log m=x+y and log n=x-y
⇒log m2 n=log m2 +log n
=2 log m+log n
=2(x+y)+x-y
=2x+2y+x-y
=3x+y

Question 16

Given that log x=m+n and log y=m-n, express the value of $\log \left(\frac{10 x}{y^{2}}\right)$ in terms of m and n
Sol :
Given that log n=m+n and log y=m-n
Then $\log \left(\frac{10 x}{y^{2}}\right)=\log 10 x-\log y^{2}$
$=\log 10+\log x-2 \log y$
=1+log x-2 log y
=1+(m+n)-2(m-n)
=1+m+n-2m+2n
=1-m+3n

Question 17

If $\frac{\log x}{2}=\frac{\log y}{3},$ find the value of $\frac{y^{4}}{x^{6}}$

Sol :

⇒$\frac{\log x}{2}=\frac{\log y}{3}$

⇒3 log x=2log y

⇒$\log x^{3}=\log y^{2}$

⇒$x^{3}=y^{2}$

Squaring both sides, we get

⇒$x^{6}=y^{4}$

⇒$y^{4}=x^{6}$

⇒$\frac{y^{4}}{x^{6}}=1$


Question 18

Solve for x:

(i) log x+log 5=2 log 3

Sol :

⇒log x+log 5=2 log 3

⇒log x=2 log 3- log 5 

⇒$\log x=\log (3)^{2}-\log 5$

⇒log x=log 9-log 5

⇒$\log x=\log \left(\frac{9}{5}\right)$

∴$x=\frac{9}{5}$


(ii) $\log _{3} x-\log _{3} 2=1$

Sol :

⇒$\log _{3} x-\log _{3} 2=1$ 

⇒$\log _{3} x=\log _{3} 2+1$

⇒$\log _{y} x=\log _{3} 2+\log _{3} 3 ~\left(\because \log _{3} 3=1\right)$

⇒$\log _{3} x=\log _{3}(2 \times 3) $

⇒$ \log _{3} x=\log _{3} 6$

∴x=6


(iii) $x=\frac{\log 125}{\log 25}$

Sol :

⇒$x=\frac{\log (5)^{3}}{\log (5)^{2}}$$x=\frac{\log (5)^{3}}{\log (5)^{2}}$

⇒$x=\frac{3 \log 5}{2 \log 5}=\frac{3}{2}$

∴$x=\frac{3}{2}$


(iv) $\frac{\log 8}{\log 2} \times \frac{\log 3}{\log \sqrt{3}}=2 \log x$

Sol :

⇒$\frac{\log (2)^{3}}{\log 2} \times \frac{\log 3}{\log (3)^{\frac{1}{2}}}=2 \log x$

⇒$\frac{3 \log 2}{\log 2} \times \frac{\log 3}{\frac{1}{2} \log 3}=2 \log x$

⇒$3(1) \times \frac{1}{\left(\frac{1}{2}\right)}=2 \log x $

⇒$3 \times \frac{2}{1}=2 \log x$

⇒2 log x=6

⇒$\log x=\frac{6}{2}$

⇒log x=3

⇒$x=(10)^{3}$

⇒x=1000


Question 19

Given $2 \log _{10} x+1=\log _{10} 250$, find

(i) x

(ii) $\log _{10} 2 x$

Sol :
(i) $2 \log _{10} x+1=\log _{10} 250$

⇒$\log _{10} x^{2}+1=\log _{10} 250$  $\left[\log _{a} m^{n}=n \log m\right]$

⇒$\log _{10} x^{2} \times 10=\log _{10} 250$ $\left[\log _{10} 10=1\right]$

⇒$\log _{10} x^{2} \times \log _{10} 10=\log _{10} 250$

⇒$x^{2} \times 10=250$

⇒$x^{2}=\frac{250}{10}$

⇒$x^{2}=25$

⇒$(x)^{2}=(5)^{2}$

∴x=5


(ii) x=5 (proved in (i) above)

⇒$\log _{10} 2 x=\log _{10} 2 \times 5$ [Putting x=5]

⇒$=\log _{10} 10=1$ $\left[\log _{10} 10=1\right]$


Question 20

If $\frac{\log x}{\log 5}=\frac{\log y^{2}}{\log 2}=\frac{\log 9}{\log \frac{1}{3}},$ find x and y.

Sol :

⇒$\frac{\log x}{\log 5}=\frac{\log y^{2}}{\log 2}=\frac{\log 9}{\log \frac{1}{3}}$

Taking first and third terms,

⇒$\frac{\log x}{\log 5}=\frac{\log 9}{\log \frac{1}{3}}$

⇒$\log x=\frac{\log 9}{\log \frac{1}{3}} \times \log 5$

⇒$\log x=\frac{3 \log (3 \times 3)}{\log 1-\log 3} \times \log 5$

⇒$\log x=\frac{\log (3)^{2}}{0-\log 3} \times \log 5$ [log 1=0]

⇒$\log x=\frac{2 \log 3}{-\log 3} \times \log 5$

⇒$\log x=\frac{2 \log 3}{\log 3} \times \log 5$

⇒log x=-2(1)×log 5

⇒log x=-2log 5

⇒log x=log (5)-2

⇒x=(5)-2

⇒$x=\frac{1}{(5)^{2}}$

⇒$x=\frac{1}{25}$

taking second and third terms,

⇒$\frac{\log y^{2}}{\log 2}=\frac{\log 9}{\log \left(\frac{1}{3}\right)}$

⇒$\log y^{2}=\frac{\log 9}{\log \left(\frac{1}{3}\right)} \times \log 2$

⇒$\log y^{2}=\frac{\log (3)^{2}}{\log 1-\log 3} \times \log 2$

⇒$\log y^{2}=\frac{2 \log 3}{0-\log 3} \times \log 2$ [log 1=0]

⇒$\log y^{2}=\frac{2 \log 3}{-\log 3} \times \log 2$

⇒$\log y^{2}=\frac{-2 \log 3}{\log 3} \times \log 2$

⇒$\log y^{2}=-2 \times \log 2 $

⇒$\log y^{2}=\log (2)^{-2}$

⇒$y^{2}=(2)^{-2}$

⇒$y=(2)^{-2 \times \frac{1}{2}}$

⇒$y=(2)^{-1}$

⇒$y=\frac{1}{2}$


Question 21

Prove the following :

(i) $3^{\log 4}=4^{\log 3}$

Sol :

⇒$3^{\log 4}=4^{\log 3}$ is true

if $\log 3^{\log 4}=\log 4^{\log 3}$ (Taking log both sides)

if log 4.log 3= log 3.log 4

if $\log _{2} 2 \cdot \log 3=\log 3 \cdot \log 2^{2}$

if 2log 2×log 3=log 3×2log 2

if 2 log 2 log 3=2 log 2 log 3

Which is true

Hence proved


(ii) $27^{\log 2}=8^{\log 3}$

Sol :
⇒ if $\log 27^{\log 2}=\log 8^{\log 3}$ (Taking log both sides)

⇒if log 2 log 27=log 3 log 8

⇒ if $\log 2 \log 3^{3}=\log 3 \log 2^{3}$

⇒if log 2.3 log 3=log 3.3 log 2

⇒if 3 log 2.log 3=3.log 2 log 3

Which is true

Hence proved


Question 22

Solve the following equations :

(i) log (2x + 3) = log 7

⇒2x+3=7

⇒2x=7-3

⇒2x=4

⇒$x=\frac{4}{2}$

∴x=2


(ii) log (x +1) + log (x – 1) = log 24

⇒log(x+1)(x-1)=log 24

⇒$\log \left(x^{2}-1\right)=\log 24$

⇒$x^{2}-1=24$

⇒$x^{2}=24+1$

⇒$x^{2}=25$

⇒$x^{2}=(5)^{2}$

∴$x^{2}=5$


(iii) log (10x + 5) – log (x – 4) = 2

⇒$\log \frac{(10 x+5)}{(x-4)}=2(\log 10)$ [∴log 10=1]

⇒$\log \frac{10 x+5}{x-4}=\log (10)^{2}$

⇒$\log \left(\frac{10 x+5}{x-4}\right)=\log 100$

⇒$\frac{10 x+5}{(x-4)}=100$

⇒10x+5=100(x-4)

⇒10x+5=100x-400

⇒10x-100x=-400-5

⇒-90x=-405

⇒$x=\frac{-405}{-90}$

⇒$x=\frac{405}{90}=\frac{81}{18}=\frac{9}{2}$

∴x=4.5


(iv) $\log _{10} 5+\log _{10}(5 x+1)=\log _{10}(x+5)+1$

⇒$\log _{10} 5 \times(5 x+1)=\log _{10}(x+5)+\log _{10} 10$ $\left[\log _{10} 10=1\right]$

⇒$\log _{10} 5(5 x+1)=\log _{10}[10 \times(x+5)]$

⇒5(5x+1)=10(x+5)

⇒25x+5=10x+5

⇒25x-10x=50-5

⇒15x=45

⇒$x=\frac{45}{15}$

∴x=3


(v) log (4y – 3) = log (2y + 1) – log3

⇒$\log 4 y-3=\log \frac{(2 y+1)}{3}$

⇒$4 y-3=\frac{2 y+1}{3}$

⇒3(4y-3)=2y+1

⇒12y-9=2y+1

⇒12y-2y=1+9

⇒10y=10

⇒$y=\frac{10}{10}=1$

∴y=1


(vi) $\log _{10}(x+2)+\log _{10}(x-2)=\log _{10} 3+3 \log _{10} 4$

⇒$\log _{10}(x+2)(x-2)=\log _{10} 3+\log _{10}(4)^{3}$

⇒$\log _{10}\left(x^{2}-2^{2}\right)=\log _{10} 3+\log _{10}(4 \times 4 \times 4)$

⇒$\log _{10}\left(x^{2}-4\right)=\log _{10} 3+\log _{10} 64$

⇒$\log _{10}\left(x^{2}-4\right)=\log _{10} 3 \times 64$

⇒$x^{2}-4=3 \times 64$

⇒$x^{2}-4=192$

⇒$x^{2}=192+4$

⇒$x^{2}=196$

⇒$x^{2}=(14)^{2}$

∴x=14

Hence proved


(vii) log(3x + 2) + log(3x – 2) = 5 log 2.

⇒$\log (3 x+2)(3 x-2)=\log 2^{5}$
⇒$\log \left(9 x^{2}-4\right)=\log 32$
Comparing both sides
⇒$9 x^{2}-4=32$
⇒$9x^{2}=32+4=36$
⇒$x^{2}=\frac{36}{9}=4=(\pm 2)^{2}$
∴x=±2
x=2

Question 23

Solve for x :

$\log _{3}(x+1)-1=3+\log _{3}(x-1)$

Sol :

⇒$\log _{3}(x+1)-1=3+\log _{3}(x-1)$

⇒$\log _{3}(x+1)-3 \log (x-1)=3+1$

⇒$\log _{3} \frac{x+1}{x-1}=4=4 \times 1=4 \log _{3} 3$ $\left(\because \log _{a} a=1\right)$

⇒$\log _{3} \frac{x+1}{x-1}=\log _{3} 3^{4}=\log _{3} 81$

∴$\frac{x+1}{x-1}=\frac{81}{1}$

⇒81x-81=x+1

⇒81x-x=1+81

⇒80x=82

∴$x=\frac{82}{80}=\frac{41}{40}=1 \frac{1}{40}$


Question 24

Solve for $x: 5^{\log x}+3^{\log x}=3^{\log x+1}-5^{\log x-1}$

Sol :

⇒$5^{\log x}+3^{\log x}=3^{\log x+1}-5^{\log x-1}$

⇒$5^{\log x}+3^{\log x}=3^{\log x} \cdot 3^{1}-5^{\log x} \cdot 5^{-1}$

⇒$5^{\log x}+3^{\log x}=3.3^{\log x}-\frac{1}{5} \cdot 5^{\log x}$

⇒$5^{\log x}+\frac{1}{5} \cdot 5^{\log x}=3.3^{\log x}-3^{\log x}$

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

⇒$ \frac{6}{5}\left(5^{\log x}\right)=2 \times 3^{\log x}$

⇒$\frac{5^{\log x}}{3^{\log x}}=\frac{2 \times 5}{6}=\left(\frac{5}{3}\right)^{1}$

⇒$\left(\frac{5}{3}\right)^{\log x}=\left(\frac{5}{3}\right)^{1}$

Comparing, we get

⇒log x=1=log 10

∴x=10


Question 25

If $\log \left(\frac{x-y}{2}\right)=\frac{1}{2}(\log x+\log y),$ prove that $x^{2}+y^{2}=6 x y$

Sol :

⇒$\log \left(\frac{x-y}{2}\right) \doteq \frac{1}{2}(\log x+\log y)$

⇒$\log \left(\frac{x-y}{2}\right)=\frac{1}{2} \log x y$ [∵log m+log n=log mn]

⇒$\log \left(\frac{x-y}{2}\right)=\log (x y)^{\frac{1}{2}}$

⇒$ \Rightarrow \frac{x y}{2}=(x y)^{\frac{1}{2}}$

⇒Squaring both sides, we get
⇒$\left(\frac{x-y}{2}\right)^{2}=\left[(x y)^{\frac{1}{2}}\right]^{2}$
⇒$\frac{(x-y)^{2}}{4}=(x y)^{\frac{1}{2} \times 2}$
⇒$(x-y)^{2}=4 \times x y$
⇒$x^{2}+y^{2}-2 x y=4 x y$
⇒$\left[\because(A-B)^{2}=A^{2}+B^{2}-2 A B\right]$
⇒$x^{2}+y^{2}=4 x y+2 x y$
⇒$x^{2}+y^{2}=6 x y$

Question 26

If $x^{2}+y^{2}=23 x y,$ Prove that $\log \frac{x+y}{5}=\frac{1}{2}(\log x+\log y)$

Sol :

⇒Given $x^{2}+y^{2}=23 x y$

⇒$x^{2}+y^{2}=25 x y-2 x y$

⇒$(x)^{2}+(y)^{2}+2 \times x \times y=25 x y$

⇒$(x+y)^{2}=25 x y$

⇒$\frac{(x+y)^{2}}{25}=x y$

Taking log on both sides, we get

⇒$\log \frac{(x+y)^{2}}{25}=\log x y$

⇒$\log \left(\frac{x+y}{5}\right)^{2}=\log x+\log y$

⇒$2 \log \frac{x+y}{5}=\log x+\log y$

⇒$\log \frac{x+y}{5}$

⇒$=\frac{1}{2} \quad(\log x+\log y)$ Proved


Question 27

If $p=\log _{10} 20$ and $q=\log _{10} 25$, find the value of x if $2 \log _{10}(x+1)=2 p-q$

Sol :
Given that $p=\log _{10} 20$ and $q=\log _{10} 25$
Then, $2 \log _{10}(x+1)=2 p-q$
Substituting the value of p and q, we get
⇒$2 \log _{10}(x+1)=2 \log _{10} 20-\log _{10} 25$
⇒$2 \log _{10}(x+1)=2 \log _{10} 20-\log _{10}(5)^{2}$
⇒$2 \log _{10}(x+1)=2 \log _{10} 20-2 \log _{10} 5$
⇒$2 \log _{10}(x+1)=2\left(\log _{10} 20-\log _{10} 5\right)$
⇒$\log _{10}(x+1)=2 \frac{\left(\log _{10} 20-\log _{10} 5\right)}{2}$
⇒$\log _{10}(x+1)=\log _{10} 20-\log _{10} 5$
⇒$\log _{10}(x+1)=\log _{10}\left(\frac{20}{5}\right)$
⇒$\log _{10}(x+1)=\log _{10} 4$
⇒x+1=4
⇒x=4-1
∴x=3

Question 28

Show that:

(i) $\frac{1}{\log _{2} 42}+\frac{1}{\log _{3} 42}+\frac{1}{\log _{7} 42}=1$

Sol :
⇒$\frac{1}{\log _{2} 42}+\frac{1}{\log _{3} 42}+\frac{1}{\log _{7} 42}=1$
L.H.S$=\frac{1}{\log _{2} 42}+\frac{1}{\log _{3} 42}+\frac{1}{\log _{7} 42}$
$\left\{\because \log _{n} m=\frac{\log _{m}}{\log _{n}}\right\}$

$=\frac{1}{\frac{\log 42}{\log _{2}}}+\frac{1}{\frac{\log 42}{\log _{3}}}+\frac{1}{\frac{\log 42}{\log _{7}}}$
$=\frac{\log _{2}}{\log 42}+\frac{\log _{3}}{\log 42}+\frac{\log _{7}}{\log 42}$
$=\frac{\log _{2}+\log _{3}+\log _{7}}{\log 42}=\frac{\log _{(2 \times 3 \times 7)}}{\log 42}$
$\left\{\begin{array}{l}\because \log _{m}+\log _{n}+\log _{p} \\ =\log _{m n p}\end{array}\right\}$
$=\frac{\log 42}{\log 42}=1$
=R.H.S

(ii) $\frac{1}{\log _{8} 36}+\frac{1}{\log _{9} 36}+\frac{1}{\log _{18} 36}=2$

Sol :
L.H.S$=\frac{1}{\log _{8} 36}+\frac{1}{\log _{9} 36}+\frac{1}{\log _{18} 36}$
$=\frac{1}{\frac{\log 36}{\log _{8}}}+\frac{1}{\frac{\log 36}{\log _{9}}}+\frac{1}{\frac{\log 36}{\log _{18}}}$
$=\frac{\log _{8}}{\log 36}+\frac{\log _{9}}{\log 36}+\frac{\log _{18}}{\log 36}$
$=\frac{\log _{8}+\log _{9}+\log _{18}}{\log 36}$
$=\frac{\log (8 \times 9 \times 18)}{\log 36}$
$=\frac{\log (36)^{2}}{\log 36}$
$=\frac{2 \log 36}{\log 36}$
=2=R.H.S

Question 29

Prove the following identities:
(i) $\frac{1}{\log _{a} a b c}+\frac{1}{\log _{b} a b c}+\frac{1}{\log _{c} a b c}=1$
Sol :
⇒$\frac{1}{\log _{a} a b c}+\frac{1}{\log _{b} a b c}+\frac{1}{\log _{c} a b c}=1$
⇒L.H.S. $=\frac{1}{\log _{a} a b c}+\frac{1}{\log _{b} a b c}+\frac{1}{\log _{c} a b c}$
⇒$=\frac{\frac{1}{\log a b c}}{\log _{a}}+\frac{1}{\frac{\log a b c}{\log _{b}}}+\frac{\frac{1}{\log a b c}}{\log _{c}}$
$\left\{\because \log _{n} m=\frac{\log _{m}}{\log _{n}}\right\}$
⇒$=\frac{\log _{a}}{\log a b c}+\frac{\log _{b}}{\log a b c}+\frac{\log _{c}}{\log a b c}$
⇒$=\frac{\log _{a}+\log _{b}+\log _{c}}{\log a b c}$
⇒$=\frac{\log a b c}{\log a b c}=1$
=R.H.S
$\left\{\begin{array}{l}\because \log m n p \\ =\log _{m}+\log _{n}+\log _{p}\end{array}\right\}$


(ii) $\log _{b} a \cdot \log _{c} b \cdot \log _{d} c=\log _{d} a$
Sol :
L.H.S$=\log _{b} a \times \log _{c} b \times \log _{d} c$

$=\frac{\log a}{\log b} \times \frac{\log b}{\log c} \times \frac{\log c}{\log d}=\frac{\log a}{\log d}$

$=\log _{d} a$
=R.H.S

Question 30

Given that $\log _{a} x=\frac{1}{\alpha}, \log _{b} x=\frac{1}{\beta}, \log _{c} x=\frac{1}{\gamma}$
Find $\log _{a b c} x$
Sol :
⇒$\log_{a} x=\frac{1}{\alpha},\log_{b}x=\frac{1}{\beta},\log_{c} x=\frac{1}{gamma}$
⇒$\log _{a} x=\frac{1}{\alpha}$
⇒$\frac{\log x}{\log _{a}}=\frac{1}{\alpha}$
⇒$\log _{a}=\alpha \log x$
⇒$\log _{b} x=\frac{1}{\beta}$
⇒$\frac{\log x}{\log _{b}}=\frac{1}{\beta} $
⇒$ \log _{b}=\beta \log x$
⇒$\log _{c} x=\frac{1}{\gamma}$
⇒$ \frac{\log x}{\log _{c}}=\frac{1}{\gamma}$
⇒ $\log _{c}=\gamma \log x$

Now $\log _{a b c} x=\frac{\log x}{\log a b c}$
⇒$=\frac{\log x}{\log a+\log b+\log c}$
⇒$=\frac{\log x}{\alpha \log x+\beta \log x+\gamma \log x}$
⇒$=\frac{\log x}{\log x(\alpha+\beta+\gamma)}=\frac{1}{\alpha+\beta+\gamma}$

Question 31

Solve for x
(i) $\log _{3} x+\log _{9} x+\log _{81} x=\frac{7}{4}$
Sol :
⇒$\frac{1}{\log _{x} 3}+\frac{1}{\log _{x} 9}+\frac{1}{\log _{x} 81}=\frac{7}{4}$
⇒$\frac{1}{\log _{x} 3^{1}}+\frac{1}{\log _{x} 3^{2}}+\frac{1}{\log _{x} 3^{4}}=\frac{7}{4}$
⇒$\frac{1}{\log _{x} 3}+\frac{1}{2 \log _{x} 3}+\frac{1}{4 \log _{x} 3}=\frac{7}{4}$
⇒$\frac{1}{\log _{x} 3}\left[1+\frac{1}{2}+\frac{1}{4}\right]=\frac{7}{4}$
⇒$\log _{x} 3 \times \frac{7}{4}=\frac{7}{4}$
⇒$\log _{x} 3=\frac{7}{4} \times \frac{4}{7}=1=\log _{3} 3$
$\left\{\because \log _{n} a=1\right\}$

Comparing, we get
∴x=3

(ii) $\log _{2} x+\log _{8} x+\log _{32} x=\frac{23}{15}$
Sol : 
⇒$\frac{1}{\log _{x} 2}+\frac{1}{\log _{x} 8}+\frac{1}{\log _{x} 32}=\frac{23}{15}$
⇒$\frac{1}{\log _{x} 2^{1}}+\frac{1}{\log _{x} 2^{3}}+\frac{1}{\log _{x} 2^{5}}=\frac{23}{15}$
⇒$\frac{1}{\log _{x} 2}+\frac{1}{3 \log _{x} 2}+\frac{1}{5 \log _{x} 2}=\frac{23}{15}$
⇒$\frac{1}{\log _{x} 2}\left[-1+\frac{1}{3}+\frac{1}{5}\right]=\frac{23}{15}$
⇒$\frac{23}{15}\left[\log _{x} 2\right]=\frac{23}{15}$
⇒$\log _{x} 2=\frac{23}{15} \times \frac{15}{23}=1=\log _{2} 2$ $\left\{\because \log _{a} a=1\right\}$

Comparing , we get
x=2

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