RBSE Solutions Class 11 Maths Chapter 9 Logarithms Exercise 9.1

RBSE Solutions for Class 11 Maths Chapter 9 Logarithms Ex 9.1

(Q. 1 to 6) Write the following in logarithm form:

Question 1. \( 2^6 = 64 \)
Answer: The logarithm form of \( 2^6 = 64 \) is \( \log_2 64 = 6 \). This conversion changes the exponential expression \( \text{base}^{\text{exponent}} = \text{result} \) into \( \log_{\text{base}}(\text{result}) = \text{exponent} \).
In simple words: If 2 raised to the power of 6 equals 64, then the logarithm with base 2 of 64 is 6.

🎯 Exam Tip: Always correctly identify the base, exponent, and result when converting between exponential and logarithmic forms to avoid errors.

Question 2. \( 10^4 = 10000 \)
Answer: The logarithm form of \( 10^4 = 10000 \) is \( \log_{10} 10000 = 4 \). The base of the logarithm always matches the base of the exponential term. Common logarithms are those with a base of 10.
In simple words: When 10 is raised to the power of 4, the result is 10000. In logarithm form, this means log base 10 of 10000 is 4.

🎯 Exam Tip: Remember that when no base is written for a logarithm, it is usually assumed to be base 10 (common logarithm).

Question 3. \( 2^{10} = 1024 \)
Answer: The logarithm form of \( 2^{10} = 1024 \) is \( \log_2 1024 = 10 \). This shows how many times the base number (2) must be multiplied by itself to get the result (1024).
In simple words: If 2 to the power of 10 is 1024, then log base 2 of 1024 is 10.

🎯 Exam Tip: Knowing common powers of small numbers (like 2, 3, 5) can help you quickly convert and solve logarithmic problems.

Question 4. \( 5^{-2} = \frac{1}{25} \)
Answer: The logarithm form of \( 5^{-2} = \frac{1}{25} \) is \( \log_5 \left(\frac{1}{25}\right) = -2 \). A negative exponent indicates that the number is the reciprocal of the base raised to the positive exponent.
In simple words: If 5 to the power of -2 equals \( \frac{1}{25} \), then log base 5 of \( \frac{1}{25} \) is -2.

🎯 Exam Tip: Remember that \( a^{-n} = \frac{1}{a^n} \). This property is crucial when dealing with negative exponents in logarithms.

Question 5. \( 10^{-3} = 0.001 \)
Answer: The logarithm form of \( 10^{-3} = 0.001 \) is \( \log_{10} 0.001 = -3 \). This form clearly shows the exponent to which 10 must be raised to obtain 0.001.
In simple words: When 10 is raised to the power of -3, the result is 0.001. So, log base 10 of 0.001 is -3.

🎯 Exam Tip: For powers of 10, the absolute value of the exponent corresponds to the number of decimal places after the '1' in numbers smaller than 1.

Question 6. \( 4^{3/2} = 8 \)
Answer: The logarithm form of \( 4^{3/2} = 8 \) is \( \log_4 8 = \frac{3}{2} \). Fractional exponents are used to represent both roots and powers. For example, \( x^{a/b} = \sqrt[b]{x^a} \).
In simple words: If 4 raised to the power of \( \frac{3}{2} \) gives 8, then the logarithm base 4 of 8 is \( \frac{3}{2} \).

🎯 Exam Tip: When converting fractional exponents, remember that the numerator indicates the power and the denominator indicates the root.

(Q. 7 to 12) Write the following in the power form:

Question 7. \( \log_5 25 = 2 \)
Answer: The exponential (power) form of \( \log_5 25 = 2 \) is \( 5^2 = 25 \). This is the reverse conversion, turning \( \log_{\text{base}}(\text{result}) = \text{exponent} \) back into \( \text{base}^{\text{exponent}} = \text{result} \).
In simple words: If log base 5 of 25 is 2, it means that 5 to the power of 2 equals 25.

🎯 Exam Tip: Always use the base of the logarithm as the base of the exponential term, and the result of the logarithm as the exponent.

Question 8. \( \log_3 729 = 6 \)
Answer: The exponential (power) form of \( \log_3 729 = 6 \) is \( 3^6 = 729 \). This directly shows the number of times 3 must be multiplied by itself to get 729.
In simple words: If log base 3 of 729 is 6, then 3 raised to the power of 6 equals 729.

🎯 Exam Tip: Knowing perfect powers helps in quickly recognizing and converting between logarithmic and exponential forms.

Question 9. \( \log_{10} 0.001 = -3 \)
Answer: The exponential (power) form of \( \log_{10} 0.001 = -3 \) is \( 10^{-3} = 0.001 \). This clearly illustrates how a negative exponent in base 10 leads to a decimal number.
In simple words: If log base 10 of 0.001 is -3, it means that 10 to the power of -3 equals 0.001.

🎯 Exam Tip: For base 10 logarithms, the exponent directly corresponds to the position of the decimal point when writing the number in scientific notation.

Question 11. \( \log_3 \left(\frac{1}{27}\right) = -3 \)
Answer: The exponential form of \( \log_3 \left(\frac{1}{27}\right) = -3 \) is \( 3^{-3} = \frac{1}{27} \). This form makes it clear that a negative exponent produces a fraction.
In simple words: If log base 3 of \( \frac{1}{27} \) is -3, then 3 raised to the power of -3 equals \( \frac{1}{27} \).

🎯 Exam Tip: Fractions in logarithmic results often indicate negative exponents when converting to exponential form.

Question 12. \( \log_{\sqrt{2}} 4 = 4 \)
Answer: The exponential (power) form of \( \log_{\sqrt{2}} 4 = 4 \) is \( (\sqrt{2})^4 = 4 \). To verify, \( (\sqrt{2})^4 = (\sqrt{2} \times \sqrt{2}) \times (\sqrt{2} \times \sqrt{2}) = 2 \times 2 = 4 \).
In simple words: If log base \( \sqrt{2} \) of 4 is 4, it means that \( \sqrt{2} \) raised to the power of 4 equals 4.

🎯 Exam Tip: Remember that raising a square root to an even power often simplifies to a whole number, e.g., \( (\sqrt{a})^4 = (a^{1/2})^4 = a^2 \).

Question 13. If \( \log_{81} x = \frac{3}{2} \), then find the value of x.
Answer:Given the equation: \( \log_{81} x = \frac{3}{2} \)
\( \implies \) Convert to exponential form: \( x = (81)^{3/2} \)
\( \implies \) Express 81 as a power of 9: \( x = (9^2)^{3/2} \)
\( \implies \) Multiply the exponents: \( x = 9^{2 \times (3/2)} = 9^3 \)
\( \implies \) Calculate the value: \( x = 729 \) So, the value of x is 729. Converting the logarithm to an exponential form is the key step to solve for x.
In simple words: Change the logarithm equation into a power equation. Since 81 is 9 squared, x becomes 9 to the power of 3, which is 729.

🎯 Exam Tip: Always use the conversion formula \( \log_b a = c \iff b^c = a \) to solve for unknown variables in logarithmic equations.

Question 14. If \( \log_{125} P = \frac{1}{6} \), then find the value of P.
Answer:Given the equation: \( \log_{125} P = \frac{1}{6} \)
\( \implies \) Convert to exponential form: \( P = (125)^{1/6} \)
\( \implies \) Express 125 as a power of 5: \( P = (5^3)^{1/6} \)
\( \implies \) Multiply the exponents: \( P = 5^{3 \times (1/6)} = 5^{1/2} \)
\( \implies \) Convert the fractional exponent to a root: \( P = \sqrt{5} \) So, the value of P is \( \sqrt{5} \). Simplifying the base to its prime factors helps in managing fractional exponents.
In simple words: Turn the logarithm equation into a power. Since 125 is 5 cubed, P becomes 5 to the power of \( \frac{1}{2} \), which is the square root of 5.

🎯 Exam Tip: When dealing with fractional exponents, it's often helpful to express the base as a power of its prime factors before multiplying exponents.

Question 15. If \( \log_4 m = 1.5 \), then find the value of m.
Answer:Given the equation: \( \log_4 m = 1.5 \)
\( \implies \) Convert to exponential form: \( m = (4)^{1.5} \)
\( \implies \) Convert the decimal exponent to a fraction: \( m = (4)^{3/2} \)
\( \implies \) Express 4 as a power of 2: \( m = (2^2)^{3/2} \)
\( \implies \) Multiply the exponents: \( m = 2^{2 \times (3/2)} = 2^3 \)
\( \implies \) Calculate the value: \( m = 8 \) So, the value of m is 8. Converting decimal exponents to fractions makes it easier to apply exponent rules.
In simple words: Change the logarithm equation into a power. Since 1.5 is \( \frac{3}{2} \) and 4 is 2 squared, m becomes 2 to the power of (2 times 1.5), which is 2 cubed, or 8.

🎯 Exam Tip: Always convert decimal exponents into their fractional form (e.g., 0.5 to \( \frac{1}{2} \), 1.5 to \( \frac{3}{2} \)) for easier calculation, especially with roots.

Question 16. Prove that \( \log_4[\log_2\{\log_2(\log_3 81)\}] = 0 \).
Answer:We need to prove that L.H.S. \( = \log_4[\log_2\{\log_2(\log_3 81)\}] = 0 \). Let's simplify the innermost logarithm first: \( \log_3 81 \) We know that \( 81 = 3^4 \). So, \( \log_3 81 = \log_3 3^4 = 4 \) (using the property \( \log_b b^x = x \)) Now, substitute this value into the next logarithm: \( \log_2(\log_3 81) = \log_2 4 \) We know that \( 4 = 2^2 \). So, \( \log_2 4 = \log_2 2^2 = 2 \) (using the property \( \log_b b^x = x \)) Next, substitute this value into the next logarithm: \( \log_2\{\log_2(\log_3 81)\} = \log_2 2 \) We know that \( \log_b b = 1 \). So, \( \log_2 2 = 1 \) Finally, substitute this value into the outermost logarithm: \( \log_4[\log_2\{\log_2(\log_3 81)\}] = \log_4 1 \) We know that \( \log_b 1 = 0 \) for any base \( b > 0, b \ne 1 \). So, \( \log_4 1 = 0 \) Therefore, L.H.S. \( = 0 \). Since L.H.S. \( = 0 \) and R.H.S. \( = 0 \), we have L.H.S. = R.H.S. Hence Proved. This step-by-step simplification confirms the proof by applying fundamental logarithmic properties from the inside out.
In simple words: We solve the nested logarithm problem from the inside out. Log base 3 of 81 is 4. Then, log base 2 of 4 is 2. After that, log base 2 of 2 is 1. Lastly, log base 4 of 1 is 0. So, it is proven that the whole expression equals 0.

🎯 Exam Tip: For problems with nested logarithms, always simplify from the innermost logarithm outwards, remembering key properties like \( \log_b b^x = x \), \( \log_b b = 1 \), and \( \log_b 1 = 0 \).