KSEEB Class 8 Maths Solutions For Chapter 9 Exponents And Powers Points To Remember
Numbers with negative exponents obey the following laws of exponents.
a)\(a^m \times a^n=a^{m+n}\)
b)\(a^m \div a^n=a^{m-n}\)
c)\(\left(a^m\right)^n=a^{m n}\)
d)\(a^m \times b^m=(a b)^m\)
e)\(a^0=1\)
f)\(\frac{a^m}{b^m}=\left(\frac{a}{b}\right)^m\)
Very small numbers can be expressed in standard form.
Use of exponents small numbers in standard form.
1) Very large and very small numbers in standard form.
2) Standard form is also called scientific notation form.
3) A number written as \(m \times 10^n\) is said to be in standard form if m is a decimal number such that \(1 \leq m<10\) and n is either a positive or a negative integer.
Read and Learn More KSEEB Solutions for Class 8 Maths
Examples: 150,000,000,000 = \(1.5 \times 10^{11}\)
Exponential notation is a powerful way to express repeated multiplication of the same number. For any non-zero rational number ‘a’ and a natural number n, the product \(a x a x a x ….. x a(n times)=a^n\).
It is known as the nth power of ‘a’ and is read as ‘a’ raised to the power ‘n’, The rational number a is called the base, and n is called exponent.
Kseeb Solutions For 8th Class Maths Chapter 9
Exponents And Powers Solutions KSEEB Class 8 Maths Chapter 9 Exercise 9.1
1. Evaluate
1) \(3^{-2}\)
Solution: \(3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
\(\left(a^{-m}=\frac{1}{a^m}\right)\)
2) \((-4)^{-2}\)
Solution: \((-4)^{-2}=\frac{1}{(-4)^2}=\frac{1}{16}\)
\(\left(a^{-m}=\frac{1}{a^m}\right)\)
3) \(\left(\frac{1}{2}\right)^{-5}=\frac{1} {(2)^{-5}}=2^5=32\)
2. simplify and express the result in power notation with a positive exponent.
1) \((-4)^5 \div(-4)^8\)
Solution: \((-4)^5 \div(-4)^8\)
= \((-4)^{5-8} \quad\left(a^m \div a^n=a^{m-n}\right)\)
= \((-4)^{-3}\)
= \(\frac{1}{(-4)^3} \quad\left(a^{-m}=\frac{1}{a^m}\right)\)
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2) \(\left(\frac{1}{2^3}\right)^2\)
Solution: \(\left(\frac{1}{2^3}\right)^2 \quad\left(\left(a^m\right)^n=a^{m m}\right)\)
=\(\left(\frac{1}{2^3}\right)^2=\frac{1}{2^6}\)
3) \((-3)^4 \times\left(\frac{5}{3}\right)^4\)
Solution: \((-3)^4 \times\left(\frac{5}{3}\right)^4\)
=\((-3)^4 \times \frac{5^4}{3^4}\)
=\((-1)^4 \times 3^4 \times \frac{5^4}{3^4} \quad\left[(a b)^m=a^m \times b^m\right]\)
=\((-1)^4 \times 5^4=5^4\)
Karnataka Board 8th Maths Chapter 9 Exponents And Powers Solutions
4) \(\left(3^{-7} \div 3^{-10}\right) \times 3^{-5}\)
Solution: \(\left(3^{-7} \div 3^{-10}\right) \times 3^{-5}\)
= \(\left(3^{-7-(-10)}\right) \times 3^{-5} \quad\left(a^m \div a^n=a^{m-n}\right)\)
= \(3^3 \times 3^{-5}\)
= \(3^{3+(-5)} \quad\left(a^m \times a^n=a^{m+n}\right)\)
= \(3^{-2}\)
= \(\frac{1}{3^2} \quad\left(a^{-m}=\frac{1}{a^m}\right)\)
5) \(2^{-3} \times(-7)^{-3}\)
Solution: \(2^{-3} \times(-7)^{-3}\)
= \(\frac{1}{2^3} \times\left(\frac{1}{(-7)^3}\right) \quad\left(a^{-m}=\frac{1}{a^m}\right)\)
= \(\frac{1}{\left[2 \times(-7)^3\right]}=\frac{1}{(-14)^3}\left(a^m \times b^m=(a b)^m\right)\)
KSEEB Class 8 Maths Key Concepts Of Exponents And Powers
3. Find the value of
1) \(\left(3^0+4^{-1}\right) \times 2^2\)
Solution: \(\left(3^0+4^{-1}\right) \times 2^2 \quad\left[a^0=1 \& a^{-m}=\frac{1}{a^m}\right]\)
=\(\left(1+\frac{1}{4}\right) \times 2^2\)
=\(\frac{5}{4} \times 4=5\)
2) \(\left(2^{-1} \times 4^{-1}\right) \div 2^{-2}\)
Solution: \(\left(2^{-1} \times 4^{-1}\right) \div 2^{-2}\)
=\(\left[2^{-1} \times\left\{(2)^2\right\}^{-1}\right] \div 2^{-2}\)
=\(\left(2^{-1} \times 2^{-2}\right) \div 2^{-2}\left[\left(a^m\right)^n=a^{m n}\right]\)
=\(2^{-1+(-2)} \div 2^{-2} \quad\left(a^m \times a^n=a^{m+n}\right)\)
=\(2^{-3} \div 2^{-2}\)
=\(2^{-3-(-2)} \quad\left(a^m \div a^n=a^{m-n}\right)\)
=\(2^{-3+2}=2^{-1} \quad\left(a^{-m}=\frac{1}{a^m}\right)\)
=\(\frac{1}{2}\)
3) \(\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}\)
Solution: \(\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}\)
=\(\left(\frac{2}{1}\right)^2+\left(\frac{3}{1}\right)^2+\left(\frac{4}{1}\right)^2 \quad\left(a^{-m}=\frac{1}{a^m}\right)\)
=\(2^2+3^2+4^2=4+9+16=29\)
4) \(\left(3^{-1}+4^{-1}+5^{-1}\right)^0\)
Solution: \(\left(3^{-1}+4^{-1}+5^{-1}\right)^0 \quad\left(a^{-m}=\frac{1}{a^m}\right)\)
=\(\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)^0\)
=\(1\left(a^0=1\right)\)
5) \(\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2\)
Solution: \(\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2\)
=\(\left\{\left(\frac{3}{2}\right)^2\right\}^2 \quad\left(a^{-m}=\frac{1}{a^m}\right)\)
=\(\left\{\frac{3^2}{(-2)^2}\right\}^2\)
=\(\left(\frac{9}{4}\right)^2 \quad\left[\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\right]\)
=\(\frac{81}{16}\)
4. Evaluate
1. \(\frac{8^{-1} \times 5^3}{2^{-4}}\)
Solution: \(\frac{8^{-1} \times 5^3}{2^{-4}}\)
=\(\frac{2^4 \times 5^3}{8^1} \quad\left(a^{-m}=\frac{1}{a^m}\right)\)
=\(\frac{2^4 \times 5^3}{2^3}=2^{4-3} \times 5^3 \quad\left(a^m \div a^n=a^{m-n}\right)\)
=\(2 \times 125=250\)
2) \(\left(5^{-1} \times 2^{-1}\right) \times 6^{-1}\)
Solution: \(\left(5^{-1} \times 2^{-1}\right) \times 6^{-1}\)
=\(\left(\frac{1}{5} \times \frac{1}{2}\right) \times \frac{1}{6} \quad\left(a^{-m}=\frac{1}{a^m}\right)\)
=\(\frac{1}{10} \times \frac{1}{6}=\frac{1}{60}\)
Kseeb 8th Standard Maths Chapter 9 Textbook Solutions
5. Find the value of m for which \(5^m \div 5^{-3}=5^5\)
Solution: \(5^m \div 5^{-3}=5^5\)
\(5^{m-(-3)}=5^5 \quad\left(a^m \div a^n=a^{m-n}\right)\)\(5^{m+3}=5^5\)
Since the powers have the same bases on both sides their respective exponents must be equal.
m + 3 = 5
m = 5 – 3
m = 2
KSEEB Class 8 Maths key concepts of Exponents and Powers
6. Evaluate
1) \(\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
Solution: \(\left\{\left(\frac{1}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
=\(\left\{\left(\frac{3}{1}\right)^1-\left(\frac{4}{1}\right)^1\right\}^{-1} \quad\left(a^{-m}=\frac{1}{a^m}\right)\)
=\((3-4)^{-1}=(-1)^{-1}=\frac{1}{-1}=-1\)
2) \(\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}\)
Solution: \(\left(\frac{5}{8}\right)^{-7} \times\left(\frac{8}{5}\right)^{-4}\)
=\(\frac{5^{-7}}{8^{-7}} \times \frac{8^{-4}}{5^{-4}}\)
=\(\frac{8^7}{5^7} \times \frac{5^4}{8^4} \quad\left(a^{-m}=\frac{1}{a^m}\right)\)
=\(\frac{8^{7-4}}{5^{7-4}}=\frac{8^3}{5^3}=\frac{512}{125} \quad\left(a^m \div a^n=a^{m-n}\right)\)
Class 8 Maths Chapter 9 Kseeb Important Questions And Answers
7. Simplify
1) \(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \quad(t \neq 0)\)
Solution: \(\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}\)
=\(\frac{5^2 \times t^{-4}}{5^{-3} \times 5 \times 2 \times t^{-8}}\)
=\(\frac{5^2 \times t^{-4}}{5^{-3+1} \times 2 \times t^{-8}} \quad\left(a^m \times a^n=a^{m+n}\right)\)
=\(\frac{5^2 \times t^{-4}}{5^{-2} \times 2 \times t^{-8}}\)
=\(\frac{5^{2-(-2)} t^{-4-(-8)}}{2} \quad\left(a^m \div a^n=a^{m-n}\right)\)
=\(\frac{5^4 t^4}{2}=\frac{625 t^4}{2}\)
2) \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)
Solution: \(\frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}\)
=\(\frac{3^{-5} \times(2 \times 5)^{-5} \times 5^3}{5^{-7} \times 6^{-5}}\)
=\(\frac{3^{-5} \times(2 \times 5)^{-5} \times 5^3}{5^{-7} \times(2 \times 3)^{-5}}\)
=\(\frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^3}{5^{-7} \times 2^{-5} \times 3^{-5}}\)
=\(3^{-5-(-5)} \times 2^{-5-(-5)} \times 5^{-5+3-(-7)}\)
=\(3^0 \times 2^0 \times 5^5 \quad\left(a^m \div a^n=a^{m-n}\right)\)
=\(5^5 \quad\left(a^0=1\right)\)
KSEEB Class 8 Maths Chapter 9 Exponents And Powers Exercise 9.2
1. Express the following numbers in standard form.
1) 0.0000000000085
Solution: \(8.5 \times 10^{-12}\)
2) 0.00000000000942
Solution: \(9.42 \times 10^{-12}\)
3) 6020000000000000
Solution: \(6.02 \times 10^{15}\)
4) 0.00000000837
Solution: [/latex]3.186 \times \times 10^{10}[/latex]
2. Express the following numbers in decimal form.
1) \(3.02 \times 10^{-6}\)
Solution: 0.00000302
2) \(4.5 \times 10^4\)
Solution: 45000
3) \(3 \times 10^{-8}\)
Solution: 0.00000003
4) \(1.0001 \times 10^9\)
Solution: 1000100000
5) \(5.8 \times 10^{12}\)
Solution: 5800000000000
6) \(3.61492 \times 10^6\)
Solution: 3614920
Exponents And Powers Class 8 Kseeb Notes Pdf
3. Express the number appearing in the following statements in standard form.
1) 1 micron is equal to \(\frac{1}{1000000} m\)
Solution: \(\frac{1}{1000000} m=1 \times 10^{-6} m\)
2) The charge of an electron is 0.000,000,000,000,000,000,16 coulomb
Solution: 0.000,000,000,000,000,000,16
=\(1.6 \times 10^{-19} \text { coulomb }\)
3) Size of a bacteria is 0.0000005m
Solution: \(0.0000005=5 \times 10^{-7} \mathrm{~m}\)
4) The size of a plant cell is 0.00001275m
Solution: \(0.00001275 \mathrm{~m}=1.275 \times 10^{-5} \mathrm{~m}\)
5) The thickness of a thick paper is 0.07mm
Solution: \(0.07 \mathrm{~mm}=7 \times 10^{-2} \mathrm{~mm}\)
4. In a stack there are 5 books each of thickness 20mm and 5 paper sheets each of thickness 0.016mm. What is the total thickness of the stack?
Solution: Thickness of each book = 20mm
Hence, thickness of 5 books = (5 x 20)mm
= 100mm
The thickness of each paper sheet = 0.016mm
Hence, thickness of 5 paper sheets = (5 x 0.016)mm = 0.080mm
Total Thickness of the stack = Thickness of 5 books + Thickness of 5 paper sheets
= (100 + 0.080)mm
= 100.08mm
= \(1.0008 x 10^2mm\)
KSEEB Maths Class 8 Chapter 9 Exponents And Powers Additional Problems
1. Find the value of \(\frac{1}{4^{-2}}\)
Solution: \(\frac{1}{4^{-2}}=4^2=4 \times 4=16 \quad\left(\frac{1}{a^{-m}}=a^m\right)\)
2. If \(x=\left(\frac{3}{7}\right)^{-3} find x^{-2}\)
Solution: Given \(x=\left(\frac{3}{7}\right)^{-3}\)
\(x^{-2}=\left[\left(\frac{3}{7}\right)^{-3}\right]^{-2} \quad\left[\left(a^m\right)^n=a^{m n}\right]\)=\(\left(\frac{3}{7}\right)^6\)
\(x^{-2}=\frac{729}{117649}\)3. simplify
\(\left[\left\{\left(\frac{-1}{2}\right)^2\right\}^{-2}\right]^{-1}\)
Solution: \(\left[\left\{\left(\frac{-1}{2}\right)^2\right\}^{-2}\right]^{-1}=\left(\frac{-1}{2}\right)^{2 x-2 x-1}\)
=\(\left(\frac{-1}{2}\right)^4 \quad\left(\left(a^m\right)^n=a^{m n}\right)\)
=\(\frac{(-1)^4}{(2)^4}=\frac{1}{16}\)
4. Find the value of x for the expression \(3^{5 x-1} \div 27=3^{-5}\)
Solution: \(3^{5 x-1} \div 27=3^{-5}\)
=\(3^{5 x-1} \div 3^3=3^{-5}\)
\(\Rightarrow \frac{3^{5 x-1}}{3^3}=3^{-5}\)
\(\Rightarrow 3^{5 x-1-3}=3^{-5}\)
\(\Rightarrow 3^{5 x-4}=3^{-5}\) (Base are equal)
5x – 4 = -5
∴ 5x = -5+4
⇒ 5x = -1
⇒ \(x=-1 / 5\)
5. What is the value of S if 379500000 is written in the form \(\mathrm{S} \times 10^m\) with m=7?
Solution: x = 379500000
⇒ \(x=37.95 \times 10^7\)
6. What is the value of \(\left(12^{-1}+8^{-2}+7^{-1}+6^{-1}\right)^0\) ?
Solution: As we know that 0 exponent of any base equals
1. Hence \(\left(12^{-1}+8^{-2}+7^{-1}+6^{-1}\right)^0=1\)
7. What is the multiplicative inverse of \(\left(\frac{-6}{13}\right)^{-99}\)?
Solution: Let the multiplicative inverse be x
then \(\left(\frac{-6}{13}\right)^{-99} \times x=1\)
(∵1 is identity element)
\(x=\left(\frac{-6}{13}\right)^{99}\)
8. What is the expression for \(8^{-3}\) as a power with the base 2 ?
Solution: \((8)^{-3}=\left(2^3\right)^{-3}\)
=\(2^{-9}\) which is the required expression
9. Find the value of K so that \(\left(\frac{5}{3}\right)^{-2} \times\left(\frac{5}{3}\right)^{-14}=\left(\frac{5}{3}\right)^{8 k}\)
Solution: \(\left(\frac{5}{3}\right)^{-2} \times\left(\frac{5}{3}\right)^{-14}=\left(\frac{5}{3}\right)^{8 k}\)
\(\left(\frac{5}{3}\right)^{-2-14}=\left(\frac{5}{3}\right)^{8 k}\)
\(\left(\frac{5}{3}\right)^{-16}=\left(\frac{5}{3}\right)^{8 k}\)
8K = -16 (comparing the exponents)
⇒ \(k=\frac{-16}{8}=-2\)
⇒ k = -2
10. Express \(\frac{1.5 \times 10^6}{2.5 \times 10^{-4}}\) in the standard form.
Solution: \(\frac{1.5 \times 10^6}{2.5 \times 10^{-4}}=\frac{1.5^{0.50 .6}}{2.50 .5_1} \times 10^6 \times 10^4\)
\(\left(\frac{1}{a^{-m}}=a^m\right)\)=\(0.6 \times 10^{10}\left(a^m \times a^n=a^{m+n}\right)\)
=\(6.0 \times 10^9\)
11. If m=2, n= -1, then find the value of the following.
1) \(m^n-n^m\)
Solution: =\((2)^{-1}-(-1)^2\)
=\(\frac{1}{2}-1\)
=\(-\frac{1}{2}\)
2) \(m^n \div n^m\)
Solution: =\((2)^{-1} \div(-1)^2\)
=\(\frac{2^{-1}}{(-1)^2}\)
=\(\frac{2^{-1}}{1}\)
=\(\frac{1}{2} \times 1=\frac{1}{2}\)
Kseeb 8th Maths Chapter 9 Exercise Solutions Step By Step
12. Simplify
\(\left[\left(\frac{-2}{3}\right)^{-2}\right]^3 \times\left(\frac{1}{3}\right)^{-4} \times 3^{-1} \times \frac{1}{6}\)
Solution: \(\left[\left(\frac{-2}{3}\right)^{-2}\right]^3 \times\left(\frac{1}{3}\right)^{-4} \times 3^{-1} \times \frac{1}{6}\)
=\(\left(\frac{-2}{3}\right)^{-6} \times\left(3^{-1}\right)^{-4} \times 3^{-1} \times \frac{1}{3 \times 2}\)
\(\left(\begin{array}{l}\left(a^m\right)^n=a^{m n} \\
\frac{1}{a^m}=a^{-m}
\end{array}\right)\)
=\((-2)^{-6} \times 3^6 \times 3^4 \times 3^{-1} \times 3^{-1} \times 2^{-1}\)
\(\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\)
=\((-2)^{-6-1} \times 3^{6+4-1-1}\)
(∵\((-2)^m=(2)^m\), if m is even)
=\(-2^{-7} \times 3^8\)
\(\Rightarrow \frac{3^8}{2^7}\)
13. By what number should \((-8)^{-3}\) be multiplied so that the product may be equal to \((-6)^{-3}\) ?
Solution: Let the number be x
\((-8)^{-3} \times x=(-6)^{-3}\)
\(\left(\frac{1}{-8}\right)^3 \times x=\left(-\frac{1}{6}\right)^{+3}\)
=\(\left(\frac{z^4}{6_3}\right)^3\)
=\(\left(\frac{4}{3}\right)^3=\frac{64}{27}\)
∴ \(x=\frac{64}{27}\)
14. Simplify \(\left(x^{\frac{b+c}{c-a}}\right)^{\frac{1}{a-b}}\left(x^{\frac{c+a}{a-b}}\right)^{\frac{1}{b-c}}\left(x^{\frac{a+b}{b-c}}\right)^{\frac{1}{c-a}}=1v prove that
Solution: LHS = [latex]\left(x^{\frac{b+c}{c-a}}\right)^{\frac{1}{a-b}}\left(x^{\frac{c+a}{a-b}}\right)^{\frac{1}{b-c}}\left(x^{\frac{a+b}{b-c}}\right)^{\frac{1}{c-a}\)
=\(\frac{b+c}{x^{(a-b)(c-a)}}+\frac{c+a}{(a-b)(b-c)}+\frac{a+b}{(b-c)(c-a)}\)
=\(x \quad \frac{(b+c)(b-c)+(c+a)(c-a)+(a+b)(a-b)}{(a-b)(b-c)(c-a)}\)
=\(x \frac{b^2-a^2+a^2-a^2+a^2-b^2}{(a-b)(b-c)(c-a)}\)
[∵\((a+b)(a-b)=a^2 – b^2\)]
=\(\frac{0}{x^{(a-b)(b-c)(c-a)}}\)
=\(x^0\)
= 1
= RHS
Hence proved.
Karnataka Board Class 8 Maths Exponents And Powers Problems And Answers
15. By what number should \(\left(\frac{1}{2}\right)^{-1}\) be divided so that the quotient is \(\left(\frac{-5}{4}\right)^{-1}\) ?
Solution: Let the required number be x.
Then \(\left(\frac{1}{2}\right)^{-1} \div x=\left(\frac{-5}{4}\right)^{-1}\)
=\(2 \times \frac{-5}{4}\)
\(x=\frac{-5}{2}\)
Hence, the required number is \(\frac{-5}{2}\).
16. If \(6^{2 x+1} \div 36=216\), find the value of x.
Solution: Given \(6^{2 x+1} \div 36=216\)
\(\frac{6^{2 x+1}}{6^2}=6^3\)
\(6^{2 x+1}=6^5\)
⇒ 2x + 1 = 5
⇒ 2x = 4
\(x=\frac{4}{2}\)
∴ x = 2
17. Express \(\frac{216}{1000}\) in exponential notation. Express the answer in lowest terms.
Solution: \(\frac{216}{1000}=\frac{6^3}{10^3}=\left(\frac{6}{10}\right)^3=\left(\frac{3}{5}\right)^3\)
18. Find the value of \(\left(3^2\right)^3+\left(\frac{2}{3}\right)^0+3^5\)
Solution: \(\left(3^2\right)^3+\left(\frac{2}{3}\right)^0+3^5\)
=\(3^6+1+3^5 \quad\left(a^m\right)^n=a^{m n}\)
= 729 + 1 + 243
= 973
19. Evaluate \(\left\{\left(\frac{4}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
Solution: \(\left\{\left(\frac{4}{3}\right)^{-1}-\left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
=\(\left[\left(\frac{3}{4}\right)^1-\left(\frac{4}{1}\right)^1\right]^{-1}\)
=\(\left(\frac{3}{4}-\frac{4}{1}\right)^{-1}\)
=\(\left(\frac{3-16}{4}\right)^{-1}\)
=\(\left(\frac{-13}{4}\right)^{-1}\)
=\(\left(\frac{-4}{13}\right)^1\)
=\(\frac{-4}{13}\)
20. Find the value of \(m+m^m(m) \text { When } m=3 \text {. }\)
Solution: \(m+m^m(m)\)
=\(3+3^3(3)\)
= 3 + 27 x 3
= 3 + 81
= 84
How To Solve Exponents And Powers Problems In Class 8 Kseeb Maths
21. A new born bear weighs 4kg, how many kilograms might a five-year-old bear weigh if its weight increases by the power of 2 in 5 yr?
Solution: Weight of newborn bear = 4kg
Weight increases by the power of 2 in 5yrs.
Weight of bearin 5 yrs =\(4^2\) = 16kg.
22. The cells of a bacteria double itself every hour. How many cells will be there after 8h. If initially, we start with cell. Express the answer in powers.
Solution: The cell of a bacteria double itself every hour = \(1 + 1 = 2 = 2^1\)
Since, the process started with 1 cell
∴ The total number of cell in 8h
= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
=\(2^{1+1+1+1+1+1+1+1}=2^8\)
23. If a = -1, b = 2, then find the value of \(a^b+b^a\text { \& } a^b \div b^a\)
Solution: 1) Given \(a^b+b^a\)
If \(a=-1 \& b=2, \text { then }(-1)^2+(2)^{-1}\)
=\(1+\frac{1}{2}=\frac{2+1}{2}=\frac{3}{2}\)
2) \(a^b \div b^a, \text { If } a=-1 \& \mathrm{~b}=2\)
\(\text { then }(-1)^2 \div(2)^{-1}=1 \div \frac{1}{2}=\frac{1 \times 2}{1}=2\)
24. Find x, \(\left(\frac{2}{5}\right)^{2 x+6} \times\left(\frac{2}{5}\right)^3=\left(\frac{2}{5}\right)^{x+2}\)
Solution: wkt \(a^m \times a^n=a^{m+n}\)
then \(\left(\frac{2}{5}\right)^{2 x+6}\) \(\times\left(\frac{2}{5}\right)^3=\left(\frac{2}{5}\right)^{2 x+6+3}=\left(\frac{2}{5}\right)^{x+2}\)
2x + 9 = x + 2
2x – x = 2 – 9
x = -7
25. Find three machines that can be replaced with hook-up of \((x^5 machines)\).
Solution: Since \(5^2=25,5^3=125,5^4=625\)
Hence \(\left(x 5^2\right),\left(x 5^3\right) \&\left(x 5^4\right)\) machine can replace \((x^5)\) hook – up machine.