Exercise 7 C
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
⇒ $\frac{(x-y)^2}{49}=xy$
⇒$\frac{(x-y)^2}{49}=xy$
⇒$\left(\frac{x-y}{7}\right)^2=xy$
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