SChand CLASS 9 Chapter 7 Logarithms Exercise 7(B)

Exercise 7 B

Question 1

Express each sum or difference as a single logarithm.

(i) $\log 6+\log 5$
$=\log (6 \times 5)=\log 30$


(ii) $\log 12-\log 2$
$=\log (12 / 2)$ $=\log 6$


(iii) $\log _{3} 5+\log _{3} 2+\log _{3} 4$
$=\log _{3}(5 \times 2 \times 4)$ $=\log _{3} 40$


(iv) $\log _{2} 12-\log _{2} 2+\log _{2} 5$
$=\log _{2}(12 \times 5 \div 2)$ $=\log _{2} 30$

Question 2

(i) $\frac{\log _{40} 1000}{\log _{40} 100}$

$=\frac{\log _{40} 10^{3}}{\log _{40} 10^{2}}=\frac{3}{2}$

 

(ii) $\frac{\log 32}{\log 4}$

$=\frac{\log 2^{5}}{\log 2^{2}}=5 / 2$

(iii) $\log _{2} 8$

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

Question 3

(i) $\log \left(m^{2}\right)-\log m$

$=\log \left(\mathrm{m}^{2} / \mathrm{m}\right)=\log \mathrm{m}$

(ii) $\log y^{2}+\log y$

$\frac{\log y^{2}}{\log y}$

$=\frac{2 \log y}{\log y}=2$

(iii)$\log 24-\log 3$

$=\log (24 / 3)$

$=log8=$

=$\log 2^{3}$

=$3 \log 2$

(iv) $\log 32+\log 4-\log 16$

$=\log (32 \times 4 / 16)$

$=\log 8=\log 2^{3}=3 \log 2$

(v) $log 256-log1024$

$=\log \left(\frac{256}{1024}\right)$

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

(vi) $\log 256 \div \log 1024$

 $=\frac{\log 256}{\log 1024}$

$=\frac{\log 4^{4}}{\log 4^{5}}$

$=4 / 5$

Question 4

Prove that:

(i)  L.H.S =$\log 7+\log 1 / 7$

$=\log 7+\log 7^{-1}$

$=\log 7-\log 7$

$=0=R \cdot H S$

(ii)R.H.S $=3 \log 2+2 \log 3$

$=\log 2^{3}+\log 3^{2}$

$=log8+log9$

$=\log (8 \times 9)=\log 72$=L.H.S

(iii) 

$\begin{aligned} R \cdot H \cdot S &=6 \log 2+\log 7 \\ &=\log 26+\log 7 \\ &=\log 64+ \log 7 \\ &=\log (64 \times 7) \end{aligned}$

=log  $448=$ L.H.S

(iv) $\log \frac{4}{47}$+ $\log 33 / 18$+ log $22 / 21$

$\log 4 / 7+\log 11 / 6-\log 22/ 21$

$\log 4-\log 7+\log 11-\log 6-\log 22$+log21

$2 \log 2-\log 7+\log 11-\log 2-\log 3$- $\log 11-\log 2$+ $4 \log 3+\log 7$

$O=R \cdot H \cdot L$

(v) $L \cdot H \cdot S=10 log 3 \sqrt{62 / 9}$

$=\log \left(\frac{56}{9}\right)^{1 / 3}$

$\frac{1}{3}(\log 56-log9)$

$\left.\frac{1}{3}( \log 7+\log 8-2 \log 3\right)$

$=\frac{1}{3}(\log 7-2 \log 3)+\log 2$=R.H.S

(vi)  L.H.S $(\log a)^{2}-(\log b)^{2}$

$=2 \log a-2 \log b$

$=2(\log a-\log b)$

$=2 \log a / b$

R.H,S  $=\log (a b)log(a / b)$

=(log a+ log b)(log a- log b)

= $(\log a)^{2}+(\log b)^{2}$

$=2(\log a-\log b)$

$=2 \log a /b$

L.H.S $=$ R.H.S

Question 5

Solve : 

(i)$\log _{10} n+\log _{10} 5=1$

sol:$\log _{10}(5 n)=1$

$5 n=10$

$n=2$

(ii) $\log _{3} n-\log _{3} 4=2$

sol: $\log _{3}(2 / 4)=2$

$n / 4=3^{2}=9$

$n=36$

(iii) $\log _{6} n-\log _{6}(n-1)=\log _{6} 3$

Sol: $\log _{6} \frac{n}{(n-1)}=\log _{6} 3$

$\frac{n}{n-1}=3$

$n=3 n-3$

$2 n=3$

$n=3 / 2$

(iv) $2 \log x=4 \log 3$

Sol:  $\log x=10 y 3^{2}$

$x=9$

Question 6

Simplify : (Do not use tables)

(i) $\log _{10} 5+\log _{10} 2$

$=\log _{10}(5 \times 2)$

$=\log _{10} 10$

$=1$

(ii) $\log _{10} 4+\log _{10} 5+\log _{10} 2$

$=\log _{10} 20-\log _{10} 2$

$=\log _{10} 20-\log _{10} 2$

$=\log _{10}(20 / 2)$

$=\log _{10} 10$

$=1$

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

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

$=\log _{10}(25 \times 8$ /$4^{1 / 2}$

$\log _{10}(25 \times 8 / 2)$

=$\log _{10} 100$

$=\log _{10} 10^{2}$

=2