SChand CLASS 9 Chapter 7 Logarithms Exercise 7(C)

  Exercise 7 C

Question 1

If $\log 2=0.3010, \log 3=0.4771, \log 5=0.6990, \log 7=0.8451$ and $\log 11=1.0414$, find the following:

(1) $\log 6$

(2) $\log 12$

(3) $\log 15$

(4) $\log 200$

(5) $\log 36$

(6) $\log 80$

(7) $\log 2$ and $1 / 3$

(8) $\log 11^{3}$

(9) $\log (2 \text { and } 1 / 3)^{5}$

Solution:

1) $\log 6$

=$\log (3 \times 2)$

=$\log 3+\log 2$

=$0.477+0.3010$

=$0.7781$

(2) 

$\begin{aligned} & \log 12 \\=& \log (3 x 4) \end{aligned}$

=$\log 3+\log 4$

$=\log 3+2 \log 2$

$=0.4771+2(0.3010)$

$=0.4771+0.6020$

$=1.0791$

6) $\log 80$

$=\log (8 \times 10)$

$=\log _{8}+\log 10$

$=\log 2^{3}+\log 10$

$=3 \log 2+9$

$=3(0.3010)+1$

$=0.9030+1$

$=1.9030$

7) $\log 21 / 3$

=$\log 7 / 3=\log 7 \mathrm-{log} 3$

=0.8451-0.477

=0.3680

8) $\log 11^{3}$

$\begin{aligned}=3 \log 11 &=3(1.0414) \\ &=3.1242 \end{aligned}$

$9) \log (2 / 3)^{5}$

=$5 \log 7 / 3$

$=5(\log 7-\log 3)$

=$5(0.845)-0.4771$

=5(0.3680)

=1.8400

Question 10

I f $\log 6=0.7782$, find the value of $\log 36$.

Sol:$\log 6=0.7782$

log 36=log $6^{2}$

$=2(0.7782)$

=1.5564

Question 11

Given $\log _{10} 25=x, \log _{10} 75=y$, evaluate without using logarithmic tables, in terms x , y

(i) $\log _{10} 3$

Sol: $\log _{10} 75=y$

$\log _{10}(3 \times 25)=y$

$\log {10}_ 3=y-x$

(ii) $\log {10}_25$

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

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

$\log _{10} 75=y$

$\log _{10}\left(5^{2} \times 3\right)=y$

$\log _{10} 5^{2}+\log _{10} 3=y$

$\log _{10} 3$=$y-2 \log _{10} 5$

$=y-2 \times x / 2$

=y-x

Question 12

Given $\log 31.87=\mathrm{x}$, write down in terms of $\mathrm{x}$ :

(i) $\log (31.87)^{2}$

Sol:$=2 \log (31.87)$

$=2 x$

(ii) $\log _{10} 0.03187$

Sol:$=\log _{10}\left(\frac{3187}{100000}\right)$

$=\log _{10}\left(\frac{31.87}{1000}\right)$

$=\log _{10}(31.871)-\log _{10} 10^{3}$

$=x-3$

(iii) $\log _{10} \sqrt{31870}$

Sol: $=\log _{10}(31870)^{1 / 2}$

$=\frac{1}{2} \log _{10}(31870)$

$=\frac{1}{2} \log _{10}\left(\frac{31870}{1000} \times 1000\right)$

$=\frac{1}{2} \log _{10}(31.87 \times 1000)$

$=1 / 2\left\{\log _{10}(31.87)+\log _{10}(1000)\right\}$

$=\frac{1}{2}\left\{x+\log _{10} 10^{3}\right\}$

$=\frac{1}{2}(x+3)=\frac{(x+3)}{2}$

Question 13

(i) $\log _{10}(x+1)+\log _{10}(x-1)=\log _{10} 11+2 \log _{10} 3$

Sol: $\log _{10}\{(x+1) \times(x-1)\}=\log _{10} 11+\log _{10} 3$

$\log _{10}\left(x^{2}-1\right)=\log _{10}(11 \times 9)$

$x^{2}-1=99$

$x^{2}=100$

$x=\pm 10$

(ii) $\log (10 x+5)-\log (x-4)=2$

=ar $\log \frac{10 x+5}{x-4}=2$

=$10 x+5=10^{2}(x-4)$

$10 x+5=100 x-400$

$90 x=405$

$x=\frac{405}{90}=4.5$

Question 14

(i) $2 \log _{10} x+\frac{1}{2} \log _{10} y=1$

$\begin{aligned} \frac{1}{2} \log _{10} y &=1-\log _{10} x^{2} \\ \log _{10} y^{1 / 2} &=\log _{10} 10^{10}-\log 10^{x} 2 \\ &=\log _{10}\left(10 / x^{2}\right) \end{aligned}$

$=\log _{10}\left(10 / x^{2}\right)$

$\therefore y^{1 / 2}=\frac{10}{x^{2}}=$

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

(ii) $2 \log 3-1 / 2 \log 16+\log 12$

$=\log 9-\log 4+\log 12$

$=2 \log 3-2 / log2+\log 3+2 \log 2$

$=3 \log 3$

Question 15

$\log _{10} y+2 \log _{10} x=2$

$\log _{10} y=2-2 \log _{10} x$

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

$=log10^{10^{2}}-\log _{10} x^{2}$

$=\log _{10}\left(\frac{100}{x^{2}}\right)$

$y=\frac{100}{x^{2}}$

$y=100 x^{-2}$

Question 17

$2+\frac{1}{2} \log _{10} 9-2 \log _{10} 5$

=2+$\frac{1}{2} \log _{10} 3^{2}=$-2log${ }_{10} 5$

$=2+\log _{10} 3$-$2 \log _{10} 5$

$=2+\log _{10} 3-$$\log _{10} 25$

=$2+\log _{10}(3 / 25)$

$=\log _{10} 10^{2}+\log _{10}\left(3^{2} / 25\right)$

$=\log _{10}\left(100 \times 3 / 25\right)$

=log${ }_{10} 2$

Question 18

(i) $a=\log{12}$,b=log 6 C=2logr2

=$a-b-c$

=$-\log 12-\log 6-2 \log 2^{1 / 2}$

=$\log (3 \times 4)-\log (3 \times 2)-\log 2$

$=\log 3+\log 4-\log 3-\log 2-\log 2$

=log 2log2 - 2log2

=0

(ii)$9 a-b-c$

$=9^{0}=1$

Question 19

(i) $x-y-z$

$=\log _{10} 12-\log _{4} 2 \times \log _{10} 9-\log _{10}(0.4)$

=$\log _{10} 12-\frac{\log 2}{2 \log 2} \times 2 \log _{10} 3-\log _{10}\left(\frac{4}{10}\right)$

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

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

=1

(ii)$6 x-y-z$

$=6^{1}=6$

Question 20

$P=\log _{10} 20$,$q=\log _{10} 25$

$=2 \log _{10} 20$- $2 \log _{10} 5$

$\log _{10}(x+1)=$ $\log _{10} 20=\log _{10} 5$

$=\log _{10}\left(\frac{20}{5}\right)$

$\begin{aligned} \therefore x+1 &=4 \\ x &=3 \end{aligned}$

Question 21

$3+\log _{10}$ $\left(10^{-2}\right)$

$=3-2 \log _{10} 10$

$=3-2=1$

Question 22

$\begin{aligned}&\log _{10} x=a \\&10 a=x\end{aligned} \quad 10^{b}=y$

(i) $10^{a-1}=10^{a}$.10-1

$=\frac{10a}{10}=\frac{x}{10}$

(ii)$10^{2b}$

=$10\left(10^{b}\right)^{2}$

=$(y)^{2}$

(iii) $\log _{10} P=2 a-b$

$=2 \log _{10} x-\log _{10} y$

$=\log _{10} x^{2}$-$\log _{10} y$

$=\log _{10}\left(x^{2} / y\right)$

$P=x^{2} / y$

Question 23

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

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

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

=$\log _{10}(25 x 8\2)

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

Question 24

(i) $\log _{10} 60$

$=\log _{10}(2 \times 3 \times 10)$

$=\log _{10} 2+\log _{10} 3+\log _{10} 10$

$=1+x+y$

Question 25

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

$\log _{10} x^{2}+1$=$\log _{10} 250$

$\log _{10} x^{2}$=$\log _{10} 250-log_{10} 10$

$=\log _{10}\left(\frac{250}{10}\right)=\log _{10}(25)$

$\therefore x^{2}=25$

$x=\pm 5$

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

$=\log _{10}(2 \times 5)$(We take only positive value )

$=\log _{10} 10$

=1

Question 26

(i)$\log _{10} \frac{\operatorname{10x}}{y^{2}}$

=$\log _{10} \frac{10 \times 10^{m+n} 1}{\left(10^{m-n}\right)^{2}}$

=$\log _{10} \frac{10^{m+n+1}}{10^{2 m-2 n}}$ 

=$\log _{10} 10^{m+n+1-2 m+2 n}$

$=\log _{10} 10^{3 n-m+1}$

$=(3 n-m+1)$

$\log x=m+n$

$x=10^{m+n}$

$\begin{aligned} \log y &=m-n \\ y &=10^{m-n} \end{aligned}$

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

$x=10^{-2}$

Question 27

If log $\left(\frac{p+q}{3}\right)=\frac{1}{2}=\text{(log p + log q)}$ prove that p² + q² = 7pq.

Sol :

$\log \left(\frac{p+q}{3}\right)$ $=\frac{1}{2}(\log p+\log q)$

$=\frac{1}{2} \log (p q)$

$=\log \sqrt{p q}$

$\frac{p+q}{3}$ =$\sqrt{p q}$

$(p+q)^{2}$=9pq

$p^{2}+2 p q$+$q^{2}$=9pq

$p^{2}+q^{2}$=7pq

Question 28

If $x^{2}+y^{2}=51 x y$, prove that log $\frac{1}{2}=\frac{1}{2}$(log x + log y).

Sol : 

⇒$x^{2}+y^{2}=51 x y$

⇒$(x-y)^{2}+2 x y=51 x y – 2xy$

(Subtracting 2xy)

⇒ (x – y)² = 49xy

⇒ $\frac{(x-y)^2}{49}=xy$

⇒$\frac{(x-y)^2}{49}=xy$

⇒$\left(\frac{x-y}{7}\right)^2=xy$