RBSE Solutions Class 8 Maths Chapter 3 Powers and Exponents Important Questions

Rajasthan Board RBSE Class 8 Maths Chapter 3 Powers and Exponents Additional Questions

I. Objective Type Questions

Question 1. Meaning of \( 2^5 \div 2^2 \) is
(a) \( 2^{5+2} \)
(b) \( 2^{5-2} \)
(c) \( 2^{5/2} \)
(d) \( (2^5)^2 \)
Answer: (b) \( 2^{5-2} \)
In simple words: When you divide numbers with the same base, you subtract their exponents. Here, the base is 2, so you subtract 2 from 5.

🎯 Exam Tip: Remember the division rule for exponents: \( a^m \div a^n = a^{m-n} \). This rule simplifies calculations significantly.

Question 2. The value of \( 3^0 \) is
(a) 0
(b) 2
(c) 1
(d) 4
Answer: (c) 1
In simple words: Any non-zero number raised to the power of zero always equals 1. It's a fundamental rule in mathematics.

🎯 Exam Tip: A common mistake is to think \( a^0 = 0 \). Always remember that \( a^0 = 1 \) for any non-zero number \( a \).

Question 3. The power in \( {\left( -\frac {8}{3} \right) }^{24} \) is
(a) 3
(b) 8
(c) 24
(d) 12
Answer: (c) 24
In simple words: The power or exponent tells us how many times the base number is multiplied by itself. In this expression, 24 is the power applied to the base \( -\frac{8}{3} \).

🎯 Exam Tip: Identify the base and the exponent clearly. The exponent is the small number written above and to the right of the base.

Question 4. The value of \( \left(\frac {1}{ { 2 }^{-2}} \right) \) will be
(a) 2
(b) 4
(c) \( \frac{5}{2} \)
(d) -3
Answer: (b) 4
In simple words: A negative exponent in the denominator means you can move the base to the numerator and make the exponent positive. So \( 2^{-2} \) in the bottom becomes \( 2^2 \) on top, which is 4.

🎯 Exam Tip: Remember the rule for negative exponents: \( \frac{1}{a^{-n}} = a^n \). This helps to quickly simplify expressions with negative powers.

Question 6. Form of Exponent in \( 4^9 \times 4^3 \) is
(a) \( 4^3 \)
(b) 27
(c) \( 4^{12} \)
(d) \( 4^6 \)
Answer: (c) \( 4^{12} \)
In simple words: When you multiply numbers with the same base, you add their exponents. Here, the base is 4, so you add 9 and 3 together.

🎯 Exam Tip: Master the multiplication rule for exponents: \( a^m \times a^n = a^{m+n} \). This is crucial for simplifying exponential expressions.

Question 7. The value of \( (3^5 \times 3^3) \div 3^{14} \) is
(a) \( \frac{1}{3^3} \)
(b) \( \frac{1}{3^6} \)
(c) \( \frac{1}{3^{14}} \)
(d) \( \frac{1}{3^3} \)
Answer: (b) \( \frac{1}{3^6} \)
In simple words: First, add the powers of 3 when they are multiplied. Then, subtract the power of 3 when it is divided. This gives a negative power, which means the number moves to the bottom of a fraction.

🎯 Exam Tip: Combine the rules of exponents: \( (a^m \times a^n) \div a^p = a^{m+n-p} \). Be careful with the order of operations.

Question 8. The value of \( (-1)^{101} \) equal to
(a) 1
(b) - 1
(c) 0
(d) 101
Answer: (b) - 1
In simple words: When a negative number like -1 is raised to an odd power, the result is always negative. If it were an even power, the result would be positive.

🎯 Exam Tip: An odd exponent always preserves the sign of the base, while an even exponent always makes the result positive. For \( (-1)^n \), if \( n \) is odd, the result is -1; if \( n \) is even, the result is 1.

Question 9. 2000000 is expressed in standard form as
(a) \( 0.2 \times 10^5 \)
(b) \( 2.0 \times 10^6 \)
(c) \( 10.2 \times 10^6 \)
(d) \( 20 \times 10^5 \)
Answer: (b) \( 2.0 \times 10^6 \)
In simple words: Standard form means writing a number as a decimal between 1 and 10, multiplied by a power of 10. To get 2 from 2000000, you move the decimal point 6 places to the left.

🎯 Exam Tip: In standard form (scientific notation), the first number must be between 1 and 10 (inclusive of 1, exclusive of 10). The exponent of 10 indicates how many places the decimal point was moved.

II. Fill in the Blanks

Question 1. Exponential form of \( \frac {3}{4} \times \frac {3}{4} \times \frac {3}{4} \) is...........
Answer: \( \left(\frac{3}{4}\right)^3 \)
In simple words: To write a repeated multiplication in exponential form, you write the number being multiplied (the base) and count how many times it's multiplied (the exponent).

🎯 Exam Tip: The base is the entire fraction \( \frac{3}{4} \), so make sure to put it in parentheses before applying the exponent.

Question 2. If a is a non zero rational number and n is a positive integer then \( a^{-n} = \).........
Answer: \( \frac {1}{a^n} \)
In simple words: A negative exponent means you take the reciprocal of the base and make the exponent positive. It's like flipping the number and then raising it to the positive power.

🎯 Exam Tip: Always remember that \( a^{-n} \) does not make the number negative; it indicates its reciprocal. \( a^{-n} = \frac{1}{a^n} \).

Question 3. A number is called in ........ form if it is expressed in \( K \times 10^n \), where \( n \) is an integer.
Answer: standard
In simple words: This special way of writing numbers with a decimal part and a power of 10 is called standard form or scientific notation. It makes very big or very small numbers easier to read.

🎯 Exam Tip: For standard form, the value of K must always be between 1 and 10 ( \( 1 \le K < 10 \) ).

Question 4. In \( {\left(\frac { 4 }{ 5 } \right) }^{50} \), then power is ...... and base is ......
Answer: power is 50 and base is \( \frac{4}{5} \)
In simple words: In any number written as \( a^n \), 'a' is the base and 'n' is the power or exponent. The base is the number being multiplied, and the power tells how many times it is multiplied.

🎯 Exam Tip: Clearly identify which part is the base and which is the exponent. The base is the entire number being raised to a power, and the exponent is the power itself.

III. True/False Type Questions

Question 1. The standard form of 100000000000 is \( 1.0 \times 10^{11} \)
Answer: True
In simple words: To write 100,000,000,000 in standard form, you place the decimal after the first digit and count the number of places it moved. Here, it moved 11 places.

🎯 Exam Tip: When moving the decimal to the left for a large number, the exponent of 10 is positive and equals the number of places moved.

Question 2. The value of \( (- 2a)^3 \) is equal to \( 8a^3 \)
Answer: False
In simple words: When a negative number is raised to an odd power, the result stays negative. So \( (-2a)^3 \) should be \( -8a^3 \), not \( 8a^3 \).

🎯 Exam Tip: Remember that \( (-x)^n \) equals \( -x^n \) if \( n \) is odd, and \( x^n \) if \( n \) is even.

Question 3. The value of \( (2^0 + 3^0) \times 4^0 \) is equal to 2
Answer: True
In simple words: Any number to the power of zero is 1. So, \( (1 + 1) \times 1 = 2 \times 1 = 2 \).

🎯 Exam Tip: Apply the rule \( a^0 = 1 \) carefully to each term in the expression before performing addition and multiplication.

Question 4. Value \( {\left(\frac {9}{8} \right) }^{ -3 }\times { {\left(\frac {8 }{9} \right) }^{-3} \) is equal to \( \frac {8}{9} \)
Answer: False
In simple words: When you have a negative exponent, you flip the fraction and make the exponent positive. So \( \left(\frac{9}{8}\right)^{-3} \) becomes \( \left(\frac{8}{9}\right)^3 \). Then you multiply \( \left(\frac{8}{9}\right)^3 \times \left(\frac{8}{9}\right)^{-3} \), which equals \( \left(\frac{8}{9}\right)^0 = 1 \). So it's 1, not \( \frac{8}{9} \).

🎯 Exam Tip: Use the rule \( \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n \) to simplify negative exponents. Also, \( a^m \times a^{-m} = a^{m-m} = a^0 = 1 \).

IV. Very Short Answer Type Questions

Question 1. Write the exponential form of \( \frac {5}{6} \times \frac {5}{6} \times \frac {5}{6} \times \frac {5}{6} \)
Answer: \( \left(\frac{5}{6}\right)^4 \)
In simple words: The number \( \frac{5}{6} \) is being multiplied by itself 4 times. So, the base is \( \frac{5}{6} \) and the power is 4.

🎯 Exam Tip: For repeated multiplication of a fraction, enclose the entire fraction in parentheses before writing the exponent to ensure it applies to both numerator and denominator.

Question 3. Write the exponential form of \( \frac{2}{5} \times \frac{2}{5} \times \frac{2}{5} \times \frac{3}{5} \times \frac{3}{5}. \)
Answer: \( {\left(\frac {2 }{ 5 } \right) }^{ 3 }\times { \left(\frac { 3 }{ 5 } \right) }^{2} \)
In simple words: Count how many times each unique fraction is multiplied. \( \frac{2}{5} \) appears 3 times, so its power is 3. \( \frac{3}{5} \) appears 2 times, so its power is 2.

🎯 Exam Tip: When different bases are multiplied, write each base with its respective exponent. Do not combine them unless their bases or exponents are the same.

Question 4. Find the value \( {\left(\frac {1}{3} \right) }^{ 3 }\times { \left( \frac {2}{3} \right) }^{4} \)
Answer:
\( {\left(\frac {1}{ 3 } \right) }^{ 3 }\times { \left( \frac { 2 }{ 3 } \right) }^{4} \)
\( = \left( \frac{1 \times 1 \times 1}{3 \times 3 \times 3} \right) \times \left( \frac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3} \right) \)
\( = \frac{1}{27} \times \frac{16}{81} \)
\( = \frac{1 \times 16}{27 \times 81} \)
\( = \frac{16}{2187} \)
In simple words: First, calculate each fraction raised to its power. Then, multiply the resulting fractions by multiplying the numerators and the denominators. This gives the final simplified value.

🎯 Exam Tip: Remember to apply the exponent to both the numerator and the denominator of a fraction. Multiply numerators and denominators separately for the final answer.

Question 5. Write the exponential form of \( { \left[ \left( { 4 }^{ 3 }\times { 4 }^{ 4 } \right) \div { 4 }^{9} \right] }^{2} \)
Answer:
\( { \left[ \left( {4}^{3}\times { 4 }^{4} \right) \div { 4 }^{ 9 } \right] }^{2} \)
\( = [4^{3+4} \div 4^9]^2 \)
\( = [4^7 \div 4^9]^2 \)
\( = [4^{7-9}]^2 \)
\( = [4^{-2}]^2 \)
\( = 4^{-2 \times 2} \)
\( = 4^{-4} \)
\( = \frac{1}{4^4} \)
In simple words: First, inside the brackets, add the powers when multiplying bases, then subtract the powers when dividing bases. Finally, multiply the outer power with the inner power. This simplifies the expression step by step.

🎯 Exam Tip: Follow the order of operations (PEMDAS/BODMAS): Parentheses/Brackets first, then Exponents, then Multiplication/Division. Apply exponent rules consistently.

Question 6. If \( \left(\frac{a}{3}\right)^{-3} = 1 \) then find the value of a.
Answer:
Given, \( \left(\frac{a}{3}\right)^{-3} = 1 \)
To make the exponent positive, we take the reciprocal of the base:
\( \implies \left(\frac{3}{a}\right)^3 = 1 \)
We know that \( 1 \) can be written as \( 1^3 \).
\( \implies \left(\frac{3}{a}\right)^3 = 1^3 \)
Since the powers are equal, their bases must also be equal.
\( \implies \frac{3}{a} = 1 \)
\( \implies a = 3 \)
In simple words: We know that anything raised to the power of zero is 1, but here the power is -3. So we flip the fraction inside to make the power positive. Since the whole thing equals 1, the fraction inside must be 1 itself, so 'a' has to be 3.

🎯 Exam Tip: Remember that any non-zero number raised to the power of zero equals 1. Also, \( x^{-n} = \frac{1}{x^n} \), which means \( \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n \).

Question 7. If \( a = 2, b = 3 \) then find the value of \( (a^b + b^a)^{-1} \).
Answer:
Given \( a = 2, b = 3 \)
We need to find the value of \( (a^b + b^a)^{-1} \)
Substitute the values of \( a \) and \( b \):
\( = (2^3 + 3^2)^{-1} \)
Calculate the powers:
\( = ((2 \times 2 \times 2) + (3 \times 3))^{-1} \)
\( = (8 + 9)^{-1} \)
Add the numbers:
\( = (17)^{-1} \)
Use the rule for negative exponents \( x^{-1} = \frac{1}{x} \):
\( = \frac{1}{17} \)
In simple words: First, put the given numbers for 'a' and 'b' into the expression. Then, work out the powers and add them up. Finally, remember that a power of -1 means you just write 1 over that number.

🎯 Exam Tip: Pay close attention to the order of operations: exponents first, then addition, then the outer exponent. Also, recall that \( x^{-1} \) is simply the reciprocal of \( x \).

Question 8. Evaluate
(i) \( 3^2 \times 3^3 \)
(ii) \( { \left[ {\left( \frac{-1}{2} \right)}^{2} \right] }^{3} \right] }^{2} \)
Answer:
(i) \( 3^2 \times 3^3 \)
When bases are the same, we add the exponents during multiplication:
\( = 3^{2+3} \)
\( = 3^5 \)
\( = 3 \times 3 \times 3 \times 3 \times 3 \)
\( = 243 \)
(ii) \( { \left[ {\left( \left(\frac{-1}{2}\right)^2 \right)}^{3} \right] }^{2} \)
When an exponent is raised to another exponent, we multiply the exponents:
\( = \left( \frac{-1}{2} \right)^{2 \times 3 \times 2} \)
\( = \left( \frac{-1}{2} \right)^{12} \)
Since the exponent is an even number, the negative sign becomes positive:
\( = \frac{1^{12}}{2^{12}} \)
\( = \frac{1}{4096} \)
In simple words: For part (i), since the base is the same, simply add the powers together, then multiply the number by itself that many times. For part (ii), multiply all the powers together. Since the final power is even, the answer will be positive. Then, calculate 1 raised to that power and 2 raised to that power.

🎯 Exam Tip: For part (i), remember \( a^m \times a^n = a^{m+n} \). For part (ii), remember \( ( (a^m)^n )^p = a^{m \times n \times p} \) and that an even exponent makes a negative base positive.

Question 2. Prove that \( x^{a-b} \times x^{b-c} \times x^{c-a} = 1 \)
Answer:
LHS \( = x^{a-b} \times x^{b-c} \times x^{c-a} \)
When multiplying terms with the same base, add their exponents:
\( = x^{(a-b) + (b-c) + (c-a)} \)
Now, simplify the exponents:
\( = x^{a-b+b-c+c-a} \)
All the positive and negative terms cancel each other out:
\( = x^0 \)
Any non-zero number raised to the power of zero is 1:
\( = 1 \)
\( = \) RHS
Thus, the equation is proven. This shows a useful property of exponents where terms cancel out to leave zero exponent, resulting in 1.
In simple words: When you multiply numbers with the same base, you add all the little powers together. If those powers cancel each other out, you are left with the power of zero, and any number with a power of zero is 1.

🎯 Exam Tip: The key here is applying the rule \( a^m \times a^n \times a^p = a^{m+n+p} \) and knowing that \( x^0 = 1 \). Be careful with the signs when adding the exponents.

Question 3. Write in ascending order of magnitude of following: \( \sqrt[4]{3}, \sqrt[3]{2}, \sqrt[3]{4} \)
Answer:
The given numbers are \( 3^{1/4}, 2^{1/3}, 4^{1/3} \).
First, find the LCM of the denominators of the fractional exponents (4, 3, 3).
LCM of 4, 3, and 3 is 12.
Now, rewrite each exponent with a denominator of 12:
\( 3^{1/4} = 3^{3/12} = (3^3)^{1/12} = 27^{1/12} \)
\( 2^{1/3} = 2^{4/12} = (2^4)^{1/12} = 16^{1/12} \)
\( 4^{1/3} = 4^{4/12} = (4^4)^{1/12} = 256^{1/12} \)
Now, compare the bases since all exponents are \( \frac{1}{12} \):
\( 16 < 27 < 256 \)
Therefore, in ascending order, we have:
\( 16^{1/12} < 27^{1/12} < 256^{1/12} \)
\( \implies 2^{1/3} < 3^{1/4} < 4^{1/3} \)
So, the ascending order is \( \sqrt[3]{2}, \sqrt[4]{3}, \sqrt[3]{4} \). Converting the fractional powers to a common denominator helps in comparing numbers with different roots.
In simple words: To compare these numbers, we make their "root" part the same. We find a common number for 4 and 3 (which is 12). Then, we change each number so it's a number raised to the power of \( \frac{1}{12} \). After that, we just compare the new numbers we got, and that gives us the correct order from smallest to largest.

🎯 Exam Tip: To compare numbers with different fractional exponents, convert them to equivalent forms with a common denominator in their exponents. The LCM of the denominators is often the easiest common denominator to use.

Question 4. Prove that \( \frac{2^{30} + 2^{29} + 2^{28}}{2^{31} + 2^{30} - 2^{29}} = \frac{7}{10} \)
Answer:
LHS \( = \frac{2^{30} + 2^{29} + 2^{28}}{2^{31} + 2^{30} - 2^{29}} \)
Take out the smallest common power of 2 from the numerator and denominator:
Numerator: \( 2^{28} (2^2 + 2^1 + 1) \)
Denominator: \( 2^{29} (2^2 + 2^1 - 1) \)
So, the expression becomes:
\( = \frac{2^{28} (2^2 + 2^1 + 1)}{2^{29} (2^2 + 2^1 - 1)} \)
Simplify the powers of 2:
\( = \frac{1}{2^{29-28}} \times \frac{(4 + 2 + 1)}{(4 + 2 - 1)} \)
\( = \frac{1}{2^1} \times \frac{7}{5} \)
\( = \frac{1}{2} \times \frac{7}{5} \)
\( = \frac{7}{10} \)
\( = \) RHS
Thus, the equation is proven. Factoring out the lowest power simplifies the expression, making it easier to solve.
In simple words: To prove this, we take out the smallest common power of 2 from the top and bottom parts of the fraction. This leaves us with simpler numbers inside the brackets. Then we cancel out the powers of 2 and solve the rest of the numbers, which will give us \( \frac{7}{10} \).

🎯 Exam Tip: When dealing with sums of exponential terms, factor out the lowest common power of the base. This often simplifies the expression dramatically and allows for cancellation.

Question 6. Find the value of p if \( 5^{p-3} \times 3^{2p-8} = 225 \)
Answer:
Given: \( 5^{p-3} \times 3^{2p-8} = 225 \)
First, express 225 as a product of powers of its prime factors:
\( 225 = 5 \times 45 = 5 \times 5 \times 9 = 5 \times 5 \times 3 \times 3 = 5^2 \times 3^2 \)
So, the equation becomes:
\( 5^{p-3} \times 3^{2p-8} = 5^2 \times 3^2 \)
Now, compare the powers of the same bases on both sides of the equation.
For base 5:
\( p-3 = 2 \)
\( \implies p = 2 + 3 \)
\( \implies p = 5 \)
For base 3:
\( 2p-8 = 2 \)
\( \implies 2p = 2 + 8 \)
\( \implies 2p = 10 \)
\( \implies p = \frac{10}{2} \)
\( \implies p = 5 \)
Both comparisons give the same value for \( p \). Therefore, the value of \( p \) is 5. By breaking down 225, we can easily compare the exponents.
In simple words: First, break down the number 225 into its prime factors, like 5 and 3. Then, match the powers of 5 on both sides of the equation, and do the same for the powers of 3. This will give you small equations that you can solve to find 'p'.

🎯 Exam Tip: When an equation involves different bases, try to express all numbers (especially the constant) as products of powers of the same bases. Then, equate the exponents of matching bases to form simpler equations.

Question 7. If \( (-2)^{x+1} \times (-2)^3 = (-2)^5 \) then find the value of x.
Answer:
Given: \( (-2)^{x+1} \times (-2)^3 = (-2)^5 \)
When multiplying terms with the same base, add their exponents:
\( (-2)^{(x+1) + 3} = (-2)^5 \)
\( (-2)^{x+4} = (-2)^5 \)
Since the bases are the same, the exponents must also be equal:
\( x+4 = 5 \)
Solve for \( x \):
\( x = 5 - 4 \)
\( x = 1 \)
So, the value of \( x \) is 1. Equating the exponents is a direct way to solve this.
In simple words: When two numbers with the same base are multiplied, you add their little power numbers together. Here, you add \( (x+1) \) and 3. Then, because the bases are the same on both sides of the equals sign, the total power on one side must be equal to the total power on the other side. This helps you find 'x'.

🎯 Exam Tip: The fundamental rule for multiplication of exponents with the same base is \( a^m \times a^n = a^{m+n} \). If \( a^m = a^n \), then \( m=n \) when \( a \ne 0, 1, -1 \).

Question 9. Write standard form of the following numbers
(a) 10350000
(b) 0.0007305
Answer:
(a) For 10350000:
Move the decimal point to the left until there is only one non-zero digit before it. The original decimal point is at the end of the number (10350000.0).
Move the decimal 7 places to the left: \( 1.035 \)
The number of places moved is the exponent of 10. Since it's a large number, the exponent is positive.
\( 10350000 = 1.035 \times 10^7 \)
(b) For 0.0007305:
Move the decimal point to the right until there is only one non-zero digit before it.
Move the decimal 4 places to the right: \( 7.305 \)
The number of places moved is the exponent of 10. Since it's a small number, the exponent is negative.
\( 0.0007305 = 7.305 \times 10^{-4} \)
In simple words: For large numbers, move the decimal left until only one digit is in front. The count of moves is a positive power of 10. For small numbers, move the decimal right until one digit is in front. The count of moves is a negative power of 10.

🎯 Exam Tip: For standard form \( a \times 10^n \), ensure \( 1 \le a < 10 \). For numbers greater than 1, \( n \) is positive. For numbers between 0 and 1, \( n \) is negative.

Question 10. The size of a red blood cell is 0.000007 m and the size of a plant cell is 0.00001275 m. Compare.
Answer:
First, write both sizes in standard form:
Size of red blood cell \( = 0.000007 \, \text{m} = 7 \times 10^{-6} \, \text{m} \)
Size of plant cell \( = 0.00001275 \, \text{m} = 1.275 \times 10^{-5} \, \text{m} \)
To compare them, we can find their ratio:
\( \frac{\text{Size of red blood cell}}{\text{Size of a plant cell}} = \frac{7 \times 10^{-6}}{1.275 \times 10^{-5}} \)
\( = \frac{7}{1.275} \times 10^{-6 - (-5)} \)
\( = \frac{7}{1.275} \times 10^{-6 + 5} \)
\( = \frac{7}{1.275} \times 10^{-1} \)
\( = \frac{7}{1.275 \times 10} \)
\( = \frac{7}{12.75} \)
\( \approx 0.549 \)
This ratio is approximately \( \frac{1}{2} \).
Therefore, a red blood cell is approximately half the size of a plant cell. Converting to scientific notation makes it easier to compare very small numbers directly.
In simple words: First, write both cell sizes using powers of 10. Then, divide the red blood cell size by the plant cell size to see how they compare. The answer shows that a red blood cell is about half the size of a plant cell.

🎯 Exam Tip: For comparison of very small or very large numbers, always convert them to standard form first. Then, either directly compare their magnitudes or calculate their ratio for a precise comparison.

Question 11. Express in standard form
(i) 15,00,00,000
(ii) 0.0000067
Answer:
(i) For 15,00,00,000:
Move the decimal point to the left until there is only one non-zero digit before it. The original decimal point is at the end of the number (150000000.0).
Move the decimal 8 places to the left: \( 1.5 \)
Since it's a large number, the exponent of 10 is positive.
\( 15,00,00,000 = 1.5 \times 10^8 \)
(ii) For 0.0000067:
Move the decimal point to the right until there is only one non-zero digit before it.
Move the decimal 6 places to the right: \( 6.7 \)
Since it's a small number, the exponent of 10 is negative.
\( 0.0000067 = 6.7 \times 10^{-6} \)
In simple words: To write 150 million in standard form, you move the decimal point 8 places to the left to get 1.5, so it's \( 1.5 \times 10^8 \). To write a very small number like 0.0000067, you move the decimal point 6 places to the right to get 6.7, so it's \( 6.7 \times 10^{-6} \).

🎯 Exam Tip: Carefully count the number of places the decimal point is moved. Moving left implies a positive exponent for \( 10^n \), while moving right implies a negative exponent.

Question 12. If, \( (243)^{x+1} = (243)^{-5} \) then find the value of x.
Answer:
Given: \( (243)^{x+1} = (243)^{-5} \)
Since the bases are the same on both sides of the equation, the exponents must be equal.
\( x+1 = -5 \)
Solve for \( x \):
\( x = -5 - 1 \)
\( x = -6 \)
Therefore, the value of \( x \) is -6. When bases are equal in an equation, we can simply equate their powers.
In simple words: When you have the same number (the base) on both sides of an equals sign, then the small power numbers (exponents) must also be equal. So, you can just set \( x+1 \) equal to -5 and solve for 'x'.

🎯 Exam Tip: The property \( \text{If } a^m = a^n \text{ (where } a \ne 0, 1, -1\text{), then } m=n \) is very useful for solving equations involving exponents.