SEBA Class 10 Advanced Maths Chapter 5 Solutions: Application of Common Logarithm
Get free SEBA Class 10 Advanced Maths Chapter 5 Solutions for the upcoming HSLC exam. In this article, we solved all the textual exercises of Chapter 5 Application of Common Logarithm from the SEBA Class 10 New Advanced Mathematics book. We provide all solutions of Chapter 5 in a very simple way.
1. Write the characteristic of the following logarithms.
(i) log 6987
Solution: Integral part of 6987 contains 4 digits.
$\therefore$ Characteristic = $(4-1)=3$
(ii) log 256987
$\therefore$ Characteristic = $(6-1)=5$
(iii) log 0.000089
Solution: There are 4 zeros between the decimal point and the first significant digit.
$\therefore$ Characteristic = $-(4+1)=-5$
(iv) log 4.68
Solution: Integral part of 4.68 contains 1 digit.
$\therefore$ Characteristic = $(1-1)=0$
(v) log 0.7491
$\therefore$ Characteristic = $-(0+1)=-1$
(vi) log 1
$\therefore$ Characteristic = $(1-1)=0$
(vii) log 35.492
$\therefore$ Characteristic = $(2-1)=1$
(viii) log 36
$\therefore$ Characteristic = $(2-1)=1$
(ix) log 0.00305
$\therefore$ Characteristic = $-(2+1)=-3$
(x) log 1.965
$\therefore$ Characteristic = $(1-1)=0$
2. If $log~2925=3.46612$ then find values of $log~29.25$, $log~2.925$, $log~0.002925$ and $log~292500$.
$Sol^{n}$:
Given, $log~2925=3.46612$
Here, Characteristic = $3$ and Mantissa = $0.46612$
Since all the given numbers contain the digits (2925), their mantissa part will remain the same as $0.46612$.
$log~29.25=$ Characteristic + Mantissa
$=(2-1)+0.46612$
$=1.46612$
$log~2.925=$ Characteristic + Mantissa
$=(1-1)+0.46612$
$=0.46612$
$log~0.002925=$ Characteristic + Mantissa
$=-(2+1)+0.46612$
$=-3+0.46612$
$=\overline{3}.46612$
$log~292500=$ Characteristic + Mantissa
$=(6-1)+0.46612$
$=5.46612$
3. Find antilogarithm with the help of table.
(i) 3.1465
Solution:
Here, Characteristic = $3$ and Mantissa = $0.1465$
From the antilogarithmic table, the number corresponding to the mantissa $0.1465$ is $14012$.
Since the characteristic is $3$, the decimal point is placed after $(3+1)=4$ digits from the left.
$\therefore Antilog~3.1465=1401.2$
(ii) $\overline{1}.8621$
Solution:
Here, Characteristic = $-1$ and Mantissa = $0.8621$
From the antilogarithmic table, the number corresponding to the mantissa $0.8621$ is $72795$.
Since the characteristic is $-1$, $Antilog~\overline{1}.8621=0.72795$
(iii) $\overline{4}.6663$
Solution:
Here, Characteristic = $-4$ and Mantissa = $0.6663$
From the antilogarithmic table, the number corresponding to the mantissa $0.6663$ is $46377$.
Since the characteristic is $-4$, $Antilog~\overline{4}.6663=0.00046377$
(iv) $-2.7917$
Solution:
$-2.7917=-2+(-0.7917)$
$=-3+(1-0.7917)$
$=-3+0.2083$
$=\overline{3}.2083$
From the antilogarithmic table, the number corresponding to the mantissa $0.2083$ is $16155$.
Since the characteristic is $-3$, $Antilog(-2.7917)=0.0016155$
(v) 2.5591
Solution:
Here, Characteristic = $2$ and Mantissa = $0.5591$.
From the antilogarithmic table, the number corresponding to the mantissa $0.5591$ is $36232$.
Since the characteristic is $2$, the decimal point is placed after $(2+1)=3$ digits from the left.
$\therefore Antilog~2.5591=362.32$
4. Given that $\log 2 = 0.30103$, $\log 3 = 0.47712$ and $\log 5 = 0.69897$, find the values of:
(i) $\log 0.002$
$\log 0.002 $
$= \log \left(\frac{2}{1000}\right) $
$= \log 2 – \log 10^3 $
$= 0.30103 – 3 $
$= – 3 + 0.30103 $
$= \bar{3}.30103)$
(ii) $\log 25$
$\log 25$
$= \log (5^2)$
$= 2 \log 5$
$= 2 \times 0.69897$
$= 1.39794$
(iii) $\log 7.2$
$\log 7.2$
$= \log \left(\frac{72}{10}\right)$
$= \log 72 – \log 10$
$= \log (2^3 \times 3^2) – 1$
$= \log (2^3) + \log (3^2) – 1$
$= 3 \log 2 + 2 \log 3 – 1$
$= 3(0.30103) + 2(0.47712) – 1$
$= 0.90309 + 0.95424 – 1$
$= 1.85733 – 1$
$= 0.85733$
(iv) $\log 300$
$\log 300$
$= \log (3 \times 100)$
$= \log 3 + \log 100$
$= \log 3 + \log 10^2$
$= \log 3 + 2 \log 10$
$= 0.47712 + 2(1)$
$= 2.47712$
(v) $\log 0.15$
$\log 0.15$
$= \log \left(\frac{15}{100}\right)$
$= \log 15 – \log 100$
$= \log (3 \times 5) – \log 10^2$
$= \log 3 + \log 5 – 2$
$= 0.47712 + 0.69897 – 2$
$= 1.17609 – 2$
$= -2 + 1.17609$
$= \bar{1}.17609$
5. Find the number of digits of the following numbers:
(i) $2^{20}$
Let $x = 2^{20}$
$\therefore \log x = \log 2^{20}$
$= 20 \times \log 2$
$= 20 \times 0.30103$
$= 6.0206$
Here characteristic of logarithm of $x$ is 6.
$\therefore$ No. of digits in the number $2^{20}$ is $6 + 1 = 7$.
(ii) $2^{25}$
Let $x = 2^{25}$
$\therefore \log x = \log 2^{25}$
$= 25 \times \log 2$
$= 25 \times 0.30103$
$= 7.52575$
Characteristic of logarithm of $x$ is 7.
$\therefore$ No. of digits in the number $2^{25}$ is $7 + 1 = 8$.
(iii) $3^{17}$
Let $x = 3^{17}$
$\therefore \log x = \log 3^{17}$
$= 17 \times \log 3$
$= 17 \times 0.47712$
$= 8.11104$
Characteristic of logarithm of $x$ is 8.
$\therefore$ No. of digits in the number $3^{17}$ is $8 + 1 = 9$.
(iv) $5^{15}$
Let $x = 5^{15}$
$\therefore \log x = \log 5^{15}$
$= 15 \times \log 5$
$= 15 \times 0.69897$
$= 10.48455$
Characteristic of logarithm of $x$ is 10.
$\therefore$ No. of digits in the number $5^{15}$ is $10 + 1 = 11$.
(v) $6^{20}$
Let $x = 6^{20}$
$\therefore \log x = \log 6^{20}$
$= 20 \times \log 6$
$= 20 \times \log (2 \times 3)$
$= 20 \times (\log 2 + \log 3)$
$= 20 \times (0.30103 + 0.47712)$
$= 20 \times 0.77815$
$= 15.563$
Characteristic of logarithm of $x$ is 15.
$\therefore$ No. of digits in the number $6^{20}$ is $15 + 1 = 16$.
(vi) $7^{13}$
Let $x = 7^{13}$
$\therefore \log x = \log 7^{13}$
$= 13 \times \log 7$
$= 13 \times 0.84510$
$= 10.9863$
Characteristic of logarithm of $x$ is 10.
$\therefore$ No. of digits in the number $7^{13}$ is $10 + 1 = 11$.
(vii) $2^{200} \times 3^{10}$
Let $x = 2^{200} \times 3^{10}$
$\therefore \log x = \log (2^{200} \times 3^{10})$
$= \log 2^{200} + \log 3^{10}$
$= 200 \times \log 2 + 10 \times \log 3$
$= 200 \times 0.30103 + 10 \times 0.47712$
$= 60.206 + 4.7712$
$= 64.9772$
Characteristic of logarithm of $x$ is 64.
$\therefore$ No. of digits in the number $2^{200} \times 3^{10}$ is $64 + 1 = 65$.
(viii) $3^{12} \times 2^{8}$
Let $x = 3^{12} \times 2^{8}$
$\therefore \log x = \log (3^{12} \times 2^{8})$
$= \log 3^{12} + \log 2^{8}$
$= 12 \times \log 3 + 8 \times \log 2$
$= 12 \times 0.47712 + 8 \times 0.30103$
$= 5.72544 + 2.40824$
$= 8.13368$
Characteristic of logarithm of $x$ is 8.
$\therefore$ No. of digits in the number $3^{12} \times 2^{8}$ is $8 + 1 = 9$.
(ix) $2^{100}$
Let $x = 2^{100}$
$\therefore \log x = \log 2^{100}$
$= 100 \times \log 2$
$= 100 \times 0.30103$
$= 30.103$
Characteristic of logarithm of $x$ is 30.
$\therefore$ No. of digits in the number $2^{100}$ is $30 + 1 = 31$.
(x) $6^{18}$
Let $x = 6^{18}$
$\therefore \log x = \log 6^{18}$
$= 18 \times \log 6$
$= 18 \times \log (2 \times 3)$
$= 18 \times (\log 2 + \log 3)$
$= 18 \times (0.30103 + 0.47712)$
$= 18 \times 0.77815$
$= 14.0067$
Characteristic of logarithm of $x$ is 14.
$\therefore$ No. of digits in the number $6^{18}$ is $14 + 1 = 15$.
6. How many zeros are there between the decimal point and the first significant digit after decimal in the following numbers?
(i) $2^{-64}$
Let $x = 2^{-64}$
$\therefore \log x = \log 2^{-64}$
$= -64 \times \log 2$
$= -64 \times 0.30103$
$= -19.26592$
$= -19 – 0.26592$
$= -20 + (1 – 0.26592)$
$= -20 + 0.73408$
$= \bar{20}.73408$
$\therefore$ Characteristic of $\log (2^{-64})$ is $-20 = -(19 + 1)$.
$\therefore$ No. of zeros just on the right side of the decimal point of the number $2^{-64}$ is 19.
(ii) $\left(\frac{1}{2}\right)^{1000}$
Let $x = \left(\frac{1}{2}\right)^{1000} = 2^{-1000}$
$\therefore \log x = \log 2^{-1000}$
$= -1000 \times \log 2$
$= -1000 \times 0.30103$
$= -301.03$
$= -301 – 0.03$
$= -302 + (1 – 0.03)$
$= -302 + 0.97$
$= \bar{302}.97$
$\therefore$ Characteristic of $\log \left(\frac{1}{2}\right)^{1000}$ is $-302 = -(301 + 1)$.
$\therefore$ No. of zeros just on the right side of the decimal point is 301.
(iii) $3^{-6}$
Let $x = 3^{-6}$
$\therefore \log x = \log 3^{-6}$
$= -6 \times \log 3$
$= -6 \times 0.47712$
$= -2.86272$
$= -2 – 0.86272$
$= -3 + (1 – 0.86272)$
$= -3 + 0.13728$
$= \bar{3}.13728$
$\therefore$ Characteristic of $\log (3^{-6})$ is $-3 = -(2 + 1)$.
$\therefore$ No. of zeros just on the right side of the decimal point is 2.
(iv) $\left(\frac{1}{3}\right)^{100}$
Let $x = \left(\frac{1}{3}\right)^{100} = 3^{-100}$
$\therefore \log x = \log 3^{-100}$
$= -100 \times \log 3$
$= -100 \times 0.47712$
$= -47.712$
$= -47 – 0.712$
$= -48 + (1 – 0.712)$
$= -48 + 0.288$
$= \bar{48}.288$
$\therefore$ Characteristic of $\log \left(\frac{1}{3}\right)^{100}$ is $-48 = -(47 + 1)$.
$\therefore$ No. of zeros just on the right side of the decimal point is 47.
(v) $(.0016)^{20}$
Let $x = (.0016)^{20}$
$\therefore \log x = \log (0.0016)^{20}$
$= 20 \times \log (0.0016)$
$= 20 \times \log (1.6 \times 10^{-3})$
$= 20 \times [\log(1.6) + \log(10^{-3})]$
$= 20 \times [\log(1.6) – 3]$
$= 20 \times [0.20412 – 3]$
$= 20 \times \bar{3}.20412$
$= 20 \times (-3 + 0.20412)$
$= -60 + 4.08240$
$= -56 + 0.08240$
$= \bar{56}.08240$
$\therefore$ Characteristic of $\log (.0016)^{20}$ is $-56 = -(55 + 1)$.
$\therefore$ No. of zeros just on the right side of the decimal point is 55.
(vi) $3^{-7}$
Let $x = 3^{-7}$
$\therefore \log x = \log 3^{-7}$
$= -7 \times \log 3$
$= -7 \times 0.47712$
$= -3.33984$
$= -3 – 0.33984$
$= -4 + (1 – 0.33984)$
$= -4 + 0.66016$
$= \bar{4}.66016$
$\therefore$ Characteristic of $\log (3^{-7})$ is $-4 = -(3 + 1)$.
$\therefore$ No. of zeros just on the right side of the decimal point is 3.
(vii) $(16.8)^{-12}$
Let $x = (16.8)^{-12}$
$\therefore \log x = \log (16.8)^{-12}$
$= -12 \times \log (16.8)$
$= -12 \times \log \left(\frac{168}{10}\right)$
$= -12 \times [\log(168) – \log(10)]$
$= -12 \times [2.22531 – 1]$
$= -12 \times 1.22531$
$= -14.70372$
$= -14 – 0.70372$
$= -15 + (1 – 0.70372)$
$= -15 + 0.29628$
$= \bar{15}.29628$
$\therefore$ Characteristic of $\log (16.8)^{-12}$ is $-15 = -(14 + 1)$.
$\therefore$ No. of zeros just on the right side of the decimal point is 14.
(viii) $7^{-4}$
Let $x = 7^{-4}$
$\therefore \log x = \log 7^{-4}$
$= -4 \times \log 7$
$= -4 \times 0.84510$
$= -3.3804$
$= -3 – 0.3804$
$= -4 + (1 – 0.3804)$
$= -4 + 0.6196$
$= \bar{4}.6196$
$\therefore$ Characteristic of $\log (7^{-4})$ is $-4 = -(3 + 1)$.
$\therefore$ No. of zeros just on the right side of the decimal point is 3.
(ix) $3^{-21}$
Let $x = 3^{-21}$
$\therefore \log x = \log 3^{-21}$
$= -21 \times \log 3$
$= -21 \times 0.47712$
$= -10.01952$
$= -10 – 0.01952$
$= -11 + (1 – 0.01952)$
$= -11 + 0.98048$
$= \bar{11}.98048$
$\therefore$ Characteristic of $\log (3^{-21})$ is $-11 = -(10 + 1)$.
$\therefore$ No. of zeros just on the right side of the decimal point is 10.
(x) $2^{-10}$
Let $x = 2^{-10}$
$\therefore \log x = \log 2^{-10}$
$= -10 \times \log 2$
$= -10 \times 0.30103$
$= -3.0103$
$= -3 – 0.0103$
$= -4 + (1 – 0.0103)$
$= -4 + 0.9897$
$= \bar{4}.9897$
$\therefore$ Characteristic of $\log (2^{-10})$ is $-4 = -(3 + 1)$.
$\therefore$ No. of zeros just on the right side of the decimal point of the number $2^{-10}$ is 3.
7. Find the values:
(i) $(7.92)^{5}$
Let $x = (7.92)^{5}$
$\therefore \log x = \log (7.92)^{5}$
$= 5 \times \log (7.92)$
$= 5 \times (0 + 0.89873)$
$= 5 \times 0.89873$
$= 4.49365$
$\therefore x = \text{Antilog } 4.49365$
$= 31162.05$
(ii) $\sqrt[3]{0.00000165}$
Let $x = \sqrt[3]{0.00000165} = (0.00000165)^{\frac{1}{3}}$
$\therefore \log x = \log (0.00000165)^{\frac{1}{3}}$
$= \frac{1}{3} \times \log (0.00000165)$
$= \frac{1}{3} \times \bar{6}.21748$
$= \frac{1}{3} \times (-6 + 0.21748)$
$= -2 + 0.07250$
$= \bar{2}.0725$
$\therefore x = \text{Antilog } \bar{2}.0725$
$= 0.01181$
(iii) $(0.264)^{\frac{1}{5}}$
Let $x = (0.264)^{\frac{1}{5}}$
$\therefore \log x = \log (0.264)^{\frac{1}{5}}$
$= \frac{1}{5} \times \log (0.264)$
$= \frac{1}{5} \times \bar{1}.42160$
$= \frac{1}{5} \times (-1 + 0.42160)$
$= \frac{1}{5} \times (-5 + 4.42160)$
$= -1 + 0.88432$
$= \bar{1}.88432$
$\therefore x = \text{Antilog } \bar{1}.88432$
$= 0.76616$
(iv) $\sqrt[5]{\frac{(625)^{4} \times (0.32)^{8}}{5^{2} \times (3125)^{3} \times (0.00432)^{2}}}$
Solution:
Let $x = \left[\frac{(625)^{4} \times (0.32)^{8}}{5^{2} \times (3125)^{3} \times (0.00432)^{2}}\right]^{\frac{1}{5}}$
$\therefore \log x = \frac{1}{5} \log \left[\frac{(625)^{4} \times (0.32)^{8}}{5^{2} \times (3125)^{3} \times (0.00432)^{2}}\right]$
$= \frac{1}{5} [4\log 625 + 8\log 0.32 – 2\log 5 – 3\log 3125 – 2\log 0.00432]$
$= \frac{1}{5} [4(2.79588) + 8(\bar{1}.50515) – 2(0.69897) – 3(3.49485) – 2(\bar{3}.63548)]$
$= \frac{1}{5} [11.18352 + 8(-0.49485) – 1.39794 – 10.48455 – 2(-2.36452)]$
$= \frac{1}{5} [11.18352 – 3.95880 – 1.39794 – 10.48455 + 4.72904]$
$= \frac{1}{5} [15.91256 – 15.84129 + 12.00000]$
$= \frac{1}{5} [12.07127]$
$= 2.41425$
$\therefore x = \text{Antilog } 2.41425$
$\therefore x = 259.5$
(v) $\sqrt[3]{\left(\frac{125 \times 147}{21 \times 32}\right)^{2}}$
Let $x = \left(\frac{125 \times 147}{21 \times 32}\right)^{\frac{2}{3}}$
$\therefore \log x = \frac{2}{3} \log \left(\frac{125 \times 147}{21 \times 32}\right)$
$= \frac{2}{3} [\log 125 + \log 147 – \log 21 – \log 32]$
$= \frac{2}{3} [\log(5^3) + \log(3 \times 7^2) – \log(3 \times 7) – \log(2^5)]$
$= \frac{2}{3} [3\log 5 + \log 3 + 2\log 7 – (\log 3 + \log 7) – 5\log 2]$
$= \frac{2}{3} [3\log 5 + \log 7 – 5\log 2]$
$= \frac{2}{3} [3(0.69897) + 0.84510 – 5(0.30103)]$
$= \frac{2}{3} [2.09691 + 0.84510 – 1.50515]$
$= \frac{2}{3} [1.43686]$
$= 0.95791$
$\therefore x = \text{Antilog } 0.95791$
$= 9.076$
8. Show that $\sqrt[5]{349388} = 12.843$ (Up to three decimal places)
Solution:
Let $x = \sqrt[5]{349388} = (349388)^{\frac{1}{5}}$
$\therefore \log x = \log(349388)^{\frac{1}{5}}$
$= \frac{1}{5} \times \log 349388$
$= \frac{1}{5} \times (5 + 0.54331)$
$= \frac{1}{5} \times 5.54331$
$= 1.10866$
$\therefore x = \text{Antilog } 1.10866$
$= 12.843$
$\therefore \sqrt[5]{349388} = 12.843$ (Up to three decimal places) (Proved)
9. Find the position of 1st significant digit on the right side of the decimal point of the number $\frac{1}{(16.8)^{12}}$.
Solution:
Let $x = \frac{1}{(16.8)^{12}} = (16.8)^{-12}$
$\therefore \log x = \log(16.8)^{-12}$
$= -12 \times \log(16.8)$
$= -12 \times \log\left(\frac{168}{10}\right)$
$= -12 \times [\log 168 – \log 10]$
$= -12 \times [2.22531 – 1]$
$= -12 \times 1.22531$
$= -14.70372$
$= -14 – 0.70372$
$= -15 + (1 – 0.70372)$
$= -15 + 0.29628$
$= \bar{15}.29628$
$\because \log x$ has characteristic $-15$.
$\therefore$ The first significant digit is at the 15th position on the right side of the decimal point.
10. Applying logarithm, find at what rate per cent will Rs. 1,60,000 amount to Rs. 1,91,844 in 2 years at compound interest?
Solution:
Given:
Principal ($P$) = Rs. 1,60,000
Amount ($A$) = Rs. 1,91,844
Time ($n$) = 2 years
Let the rate of interest be $r\%$ per annum.
We know that:
$A = P\left(1 + \frac{r}{100}\right)^n$
$\therefore 191844 = 160000\left(1 + \frac{r}{100}\right)^2$
$\Rightarrow \left(1 + \frac{r}{100}\right)^2 = \frac{191844}{160000}$
$\Rightarrow \log\left(1 + \frac{r}{100}\right)^2 = \log\left(\frac{191844}{160000}\right)$ [Taking logarithm on both sides]
$\Rightarrow 2\log\left(1 + \frac{r}{100}\right) = \log 191844 – \log 160000$
$\Rightarrow 2\log\left(1 + \frac{r}{100}\right) = 5.28295 – 5.20412$
$\Rightarrow 2\log\left(1 + \frac{r}{100}\right) = 0.07883$
$\Rightarrow \log\left(1 + \frac{r}{100}\right) = \frac{0.07883}{2} = 0.039415$
$\therefore 1 + \frac{r}{100} = \text{Antilog } 0.039415$
$\Rightarrow 1 + \frac{r}{100} = 1.095$
$\Rightarrow \frac{r}{100} = 1.095 – 1$
$\Rightarrow \frac{r}{100} = 0.095$
$\therefore r = 0.095 \times 100 = 9.5\%$
$\therefore$ The required rate of interest is $9.5\%$.
11. With the help of logarithm find the number of years in which Rs. 216 will become Rs. 625 at 16% compound interest.
Solution:
Given: Principal ($P$) = Rs. 216
Amount ($A$) = Rs. 625
Rate of interest ($r$) = $16\%$ per annum
Let the required time be $n$ years.
We know that:
$A = P\left(1 + \frac{r}{100}\right)^n$
$\therefore 625 = 216\left(1 + \frac{16}{100}\right)^n$
$\Rightarrow \frac{625}{216} = \left(1 + 0.16\right)^n$
$\Rightarrow (1.16)^n = \frac{625}{216}$
$\Rightarrow \log(1.16)^n = \log\left(\frac{625}{216}\right)$ [Taking logarithm on both sides]
$\Rightarrow n\log 1.16 = \log 625 – \log 216$
$\Rightarrow n(0.06446) = 2.79588 – 2.33445$
$\Rightarrow n(0.06446) = 0.46143$
$\Rightarrow n = \frac{0.46143}{0.06446}$
$\Rightarrow n = 7.158 \approx 7.15 \text{ years }$
$\therefore$ The required time is $7.15\text{ years}$ (approx.)
12. What will be the interest of Rs. 1,00,000 in 3 years at 8% compound interest?
Solution:
Given: Principal ($P$) = Rs. 1,00,000
Time ($n$) = 3 years
Rate ($r$) = 8%
Compound Interest ($CI$) = ?
We know:
$A = P\left(1 + \frac{r}{100}\right)^n$
$A = 100000\left(1 + \frac{8}{100}\right)^3$
$A = 100000(1.08)^3$
$\log A = \log[100000 \times (1.08)^3]$ [Taking logarithm on both sides]
$= \log 100000 + 3 \log 1.08$
$= 5 + 3(0.03342)$
$= 5 + 0.10026$
$= 5.10026$
$\therefore A = \text{Antilog } 5.10026$
$= 125971.20$
$\therefore$ Compound Interest ($CI$) = $A – P$
$= 125971.20 – 100000$
$= \text{Rs. } 25,971.20$
$\therefore$ The required interest is Rs. 25,971.20.
13. Find the area of the triangle having sides 45cm, 57cm and 63cm.
Solution:
Let the sides of the triangle be $a = 45\text{ cm}$, $b = 57\text{ cm}$, $c = 63\text{ cm}$.
Semi-perimeter, $s = \frac{a + b + c}{2} $
$= \frac{45 + 57 + 63}{2} $
$= \frac{165}{2} = 82.5\text{ cm}$
Here:
$s – a = 82.5 – 45 = 37.5\text{ cm}$
$s – b = 82.5 – 57 = 25.5\text{ cm}$
$s – c = 82.5 – 63 = 19.5\text{ cm}$
Area of triangle, $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$
$= [82.5 \times 37.5 \times 25.5 \times 19.5]^{\frac{1}{2}}$
$\log \Delta = \frac{1}{2}[\log 82.5 + \log 37.5 + \log 25.5 + \log 19.5]$[Taking logarithm on both sides]
$= \frac{1}{2}[1.91645 + 1.57403 + 1.40654 + 1.29003]$
$= \frac{1}{2}[6.18705]$
$= 3.09353$
$\therefore \Delta = \text{Antilog } 3.09353$
$= 1240.2\text{ sq. cm}$
$\therefore$ The area of the triangle is 1240.2 sq. cm.
14. The volume of a sphere is 72.81 cubic centimetres. Find its radius.
Solution:
Let the radius of the sphere be $r\text{ cm}$.
$\therefore$ Volume of sphere, $V = \frac{4}{3}\pi r^3$
Given, $V = 72.81\text{ cubic cm}$
$\therefore 72.81 = \frac{4}{3} \times \frac{22}{7} \times r^3$
$\Rightarrow r^3 = \frac{72.81 \times 21}{88}$
$\Rightarrow \log r^3 = \log\left(\frac{72.81 \times 21}{88}\right)$[Taking logarithm on both sides]
$\Rightarrow 3\log r = \log 72.81 + \log 21 – \log 88$
$= 1.86219 + 1.32222 – 1.94448$
$= 3.18441 – 1.94448$
$= 1.23993$
$\therefore \log r = \frac{1}{3} \times 1.23993 = 0.41331$
$\therefore r = \text{Antilog } 0.41331$
$= 2.59\text{ cm}$
$\therefore$ The required radius is $2.59\text{ cm}$.
15. Solve with the help of logarithm:
(i) $10^x = 893$
$\log 10^x = \log 893$ [Taking logarithm on both sides]
$\Rightarrow x \log 10 = \log 893$
$\Rightarrow x(1) = 2.95085$
$\therefore x = 2.95085$
(ii) $18 \times 3^x = 2^{2x}$
$\Rightarrow \log(18 \times 3^x) = \log 2^{2x}$ [Taking logarithm on both sides]
$\Rightarrow \log 18 + \log 3^x = \log 2^{2x}$
$\Rightarrow \log 18 + x\log 3 = 2x\log 2$
$\Rightarrow \log 18 = 2x\log 2 – x\log 3$
$\Rightarrow \log 18 = x(2\log 2 – \log 3)$
$\Rightarrow x = \frac{\log 18}{2\log 2 – \log 3}$
$= \frac{1.25527}{2(0.30103) – 0.47712}$
$= \frac{1.25527}{0.60206 – 0.47712}$
$= \frac{1.25527}{0.12494}$
$= 10.047$
$\therefore x = 10.047$
(iii) $7^{3x+2} + 4^{x+2} = 7^{3x+1} + 2^{2x+6}$
$\Rightarrow 7^{3x} \cdot 7^2 + 4^x \cdot 4^2 = 7^{3x} \cdot 7 + 2^{2x} \cdot 2^6$
$\Rightarrow 49 \cdot 7^{3x} – 7 \cdot 7^{3x} = 64 \times 2^{2x} – 16 \times 2^{2x}$
$\Rightarrow (49 – 7) \times 7^{3x} = 2^{2x} (64 – 16)$
$\Rightarrow 42 \times 7^{3x} = 2^{2x} \times 48$
$\Rightarrow 7 \times 7^{3x} = 8 \times 2^{2x}$
$\Rightarrow 7^{3x+1} = 2^{2x+3}$
$\therefore \log 7^{3x+1} = \log 2^{2x+3}$
$\Rightarrow (3x + 1) \log 7 = (2x + 3) \log 2$
$\Rightarrow 3x \log 7 + \log 7 = 2x \log 2 + 3 \log 2$
$\Rightarrow 3x \log 7 – 2x \log 2 = 3 \log 2 – \log 7$
$\Rightarrow x(3 \log 7 – 2 \log 2) = 3 \log 2 – \log 7$
$\Rightarrow x = \frac{3 \log 2 – \log 7}{3 \log 7 – 2 \log 2}$
$= \frac{3(0.30103) – 0.84510}{3(0.84510) – 2(0.30103)}$
$= \frac{0.90309 – 0.84510}{2.53530 – 0.60206}$
$= \frac{0.05799}{1.93324}$
$\approx 0.03$
$\therefore x = 0.03$
(iv) $\log(x^2 – 3x) = 1$
$\log(x^2 – 3x) = \log 10$
$\Rightarrow x^2 – 3x = 10$
$\Rightarrow x^2 – 3x – 10 = 0$
$\Rightarrow x^2 – 5x + 2x – 10 = 0$
$\Rightarrow x(x – 5) + 2(x – 5) = 0$
$\Rightarrow (x – 5)(x + 2) = 0$
Either, $ (x – 5)= 0$ or, $ (x + 2) = 0$
$\therefore x = 5 \text{ or } x = -2$
(v) $6^{3-4x} \times 4^{x+5} = 8$
$\log(6^{3-4x} \times 4^{x+5}) = \log 8$ [Taking logarithm on both sides]
$\Rightarrow \log 6^{3-4x} + \log 4^{x+5} = \log 2^3$
$\Rightarrow (3 – 4x)\log 6 + (x + 5)\log 4 = 3\log 2$
$\Rightarrow (3 – 4x)(\log 2 + \log 3) + 2(x + 5)\log 2 = 3\log 2$
$\Rightarrow (3 – 4x)(0.77815) + (2x + 10)(0.30103) = 3(0.30103)$
$\Rightarrow 2.33445 – 3.1126x + 0.60206x + 3.01030 = 0.90309$
$\Rightarrow 5.34475 – 2.51054x = 0.90309$
$\Rightarrow 2.51054x = 5.34475 – 0.90309$
$\Rightarrow 2.51054x = 4.44166$
$\Rightarrow x = \frac{4.44166}{2.51054} \approx 1.77$
$\therefore x = 1.77$
16. Show that : $\left(\frac{21}{10}\right)^{100} > 100$
$Sol^n .:$
Let $x = \left(\frac{21}{10}\right)^{100}$
$\therefore \log x = \log\left(\frac{21}{10}\right)^{100}$
$= 100 \log\left(\frac{21}{10}\right)$
$= 100 [\log 21 – \log 10]$
$= 100 [1.32222 – 1]$
$= 100 \times 0.32222$
$= 32.222 > 2$
$\therefore \log x > 2$
$\Rightarrow \log x > \log 100$ [ Since: $\log 100 = \log 10^2 = 2$]
$\therefore x > 100$
$\therefore \left(\frac{21}{10}\right)^{100} > 100$ (Proved)