OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Exercise 7 (B)

Question 1. Express each sum or difference as a single logarithm :
(i) log 6 + log 5
(ii) log 12 – log 2
(iii) log\(_{3}\) 5 + log\(_{3}\) 2 + log\(_{3}\) 4
(iv) log\(_{2}\) 12 – log\(_{2}\) 2 + log\(_{2}\) 5
Answer:
(i) \( \log 6 + \log 5 = \log (6 \times 5) = \log 30 \)
\( \implies \) When you add logarithms with the same base, you multiply the numbers inside the log.
(ii) \( \log 12 - \log 2 = \log \left( \frac{12}{2} \right) = \log 6 \)
\( \implies \) When you subtract logarithms with the same base, you divide the numbers inside the log.
(iii) \( \log_{3} 5 + \log_{3} 2 + \log_{3} 4 = \log_{3} (5 \times 2 \times 4) = \log_{3} 40 \)
\( \implies \) When adding multiple logarithms with the same base, you multiply all the numbers together inside one log.
(iv) \( \log_{2} 12 - \log_{2} 2 + \log_{2} 5 = \log_{2} \left( \frac{12}{2} \times 5 \right) = \log_{2} (6 \times 5) = \log_{2} 30 \)
\( \implies \) Combine addition and subtraction by multiplying and dividing the numbers inside the logarithm in the correct order. Logarithms help simplify complex multiplications and divisions into simpler additions and subtractions.
In simple words: When you see plus signs between logs, multiply the numbers. When you see minus signs, divide the numbers. Keep the log sign and its base.

🎯 Exam Tip: Remember the basic rules: \( \log a + \log b = \log (ab) \) and \( \log a - \log b = \log \left(\frac{a}{b}\right) \). These are essential for simplifying logarithmic expressions.

Question 2. Simplify without using tables :
(i) \( \frac{\log_{40} 1000}{\log_{40} 100} \)
(ii) \( \frac{\log 32}{\log 4} \)
(iii) \( \log_{2} 8 \)
Answer:
(i) \( \frac{\log_{40} 1000}{\log_{40} 100} = \frac{\log_{40} (10^3)}{\log_{40} (10^2)} \)
\( \implies = \frac{3 \log_{40} 10}{2 \log_{40} 10} \)
\( \implies = \frac{3}{2} \)
\( \implies \) Use the power rule to bring exponents down, then simplify the fraction by canceling common terms.
(ii) \( \frac{\log 32}{\log 4} = \frac{\log (2^5)}{\log (2^2)} \)
\( \implies = \frac{5 \log 2}{2 \log 2} \)
\( \implies = \frac{5}{2} \)
\( \implies \) Express numbers as powers of the same base (here, base 2), then use the power rule and simplify. This method avoids needing a calculator or log tables.
(iii) \( \log_{2} 8 = \log_{2} (2^3) \)
\( \implies = 3 \log_{2} 2 \)
\( \implies = 3 \times 1 \) (since \( \log_{a} a = 1 \))
\( \implies = 3 \)
\( \implies \) Write the number as a power of the base, then apply the logarithm rules to simplify to a whole number. This is a fundamental concept in logarithms.
In simple words: Change numbers inside the log to powers of the log's base. This makes the log simple to calculate.

🎯 Exam Tip: Always look for ways to express the numbers as powers of the base of the logarithm or a common base. This allows you to use the power rule \( \log a^n = n \log a \) to simplify the expression.

Question 3. Simplify:
(i) log (m\(^{2}\)) – log m
(ii) log y\(^{2}\) ÷ log y
(iii) log 24 - log 3
(iv) log 32 + log 4 – log 16
(v) log 256 – log 1024
(vi) log 256 – log 1024
Answer:
(i) \( \log (m^2) - \log m = 2 \log m - \log m \)
\( \implies = \log m \)
\( \implies \) Bring down the exponent using the power rule, then combine like terms. Always simplify to the simplest form.
(ii) \( \log y^2 \div \log y = 2 \log y \div \log y \)
\( \implies = 2 \)
\( \implies \) Simplify by using the power rule for the numerator and then dividing the common logarithmic term. The values of y must be positive for the logarithms to be defined.
(iii) \( \log 24 - \log 3 = \log \left( \frac{24}{3} \right) \)
\( \implies = \log 8 \)
\( \implies = \log (2^3) \)
\( \implies = 3 \log 2 \)
\( \implies \) First, use the subtraction rule for logs, then express the result as a power of 2 to simplify further. Breaking numbers into prime factors helps.
(iv) \( \log 32 + \log 4 - \log 16 = \log \left( \frac{32 \times 4}{16} \right) \)
\( \implies = \log \left( \frac{128}{16} \right) \)
\( \implies = \log 8 \)
\( \implies = \log (2^3) \)
\( \implies = 3 \log 2 \)
\( \implies \) Combine addition and subtraction of logarithms into a single logarithm, then simplify the argument using powers of 2. Order of operations is important for accurate simplification.
(v) \( \log 256 - \log 1024 = \log \left( \frac{256}{1024} \right) \)
\( \implies = \log \left( \frac{1}{4} \right) \)
\( \implies = \log (4^{-1}) \)
\( \implies = \log ((2^2)^{-1}) \)
\( \implies = \log (2^{-2}) \)
\( \implies = -2 \log 2 \)
\( \implies \) Use the subtraction rule, then express the fraction as a negative power of 2, remembering that a negative exponent creates a reciprocal.
(vi) \( \frac{\log 256}{\log 1024} = \frac{\log (2^8)}{\log (2^{10})} \)
\( \implies = \frac{8 \log 2}{10 \log 2} \)
\( \implies = \frac{8}{10} \)
\( \implies = \frac{4}{5} \)
\( \implies \) Express both numbers as powers of a common base, then use the power rule for logarithms and simplify the resulting fraction. This simplifies the expression by canceling out the common logarithmic term.
In simple words: Use the rules to put all numbers into one log, or change them into powers of the same small number, then simplify. Remember that \( \log m^n = n \log m \).

🎯 Exam Tip: When simplifying, always try to express numbers as powers of a common base (like 2, 3, 5, or 10) to make it easier to apply the power rule of logarithms. Watch out for questions that might be identical or similar, like (v) and (vi) here, but apply the rules carefully for each.

Question 4. Prove that:
(i) log 7 + log \( \frac { 1 }{ 7 } \) = 0
(ii) log 72 = 3 log 2 + 2 log 3
(iii) log 448 = 6 log 2 + log 7
(iv) log \( \frac { 4 }{ 7 } \) + log \( \frac { 33 }{ 18 } \) – log \( \frac { 22 }{ 21 } \) = 0
(v) log \( \sqrt[3]{6 \frac{2}{9}} \) = \( \frac { 1 }{ 3 }(\log 2 – 2 \log 3) + \log 2 \)
(vi) (log a)\(^{2}\) – (log b)\(^{2}\) = log (ab) log \( \frac { a }{ b } \)
Answer:
(i) To prove: \( \log 7 + \log \frac{1}{7} = 0 \)
\( \text{L.H.S.} = \log 7 + \log \frac{1}{7} \)
\( \implies = \log \left( 7 \times \frac{1}{7} \right) \) (Using \( \log m + \log n = \log (mn) \))
\( \implies = \log 1 \)
\( \implies = 0 \) (Since \( \log 1 = 0 \))
\( \implies = \text{R.H.S.} \)
\( \implies \) When adding logarithms, multiply the numbers inside the log. The logarithm of 1 to any base is always 0. This rule is very useful.
(ii) To prove: \( \log 72 = 3 \log 2 + 2 \log 3 \)
\( \text{L.H.S.} = \log 72 \)
\( \implies = \log (8 \times 9) \)
\( \implies = \log (2^3 \times 3^2) \)
\( \implies = \log (2^3) + \log (3^2) \) (Using \( \log (mn) = \log m + \log n \))
\( \implies = 3 \log 2 + 2 \log 3 \) (Using \( \log m^n = n \log m \))
\( \implies = \text{R.H.S.} \)
\( \implies \) Factorize 72 into its prime factors, then use the product rule and power rule for logarithms to expand the expression. Prime factorization is a key step here.
(iii) To prove: \( \log 448 = 6 \log 2 + \log 7 \)
\( \text{L.H.S.} = \log 448 \)
\( \implies = \log (64 \times 7) \)
\( \implies = \log (2^6 \times 7) \)
\( \implies = \log (2^6) + \log 7 \) (Using \( \log (mn) = \log m + \log n \))
\( \implies = 6 \log 2 + \log 7 \) (Using \( \log m^n = n \log m \))
\( \implies = \text{R.H.S.} \)
\( \implies \) Break down 448 into its prime factors. Then, apply the logarithm rules for products and powers to expand the left side to match the right side. This shows how prime factors relate to log expressions.
(iv) To prove: \( \log \frac{4}{7} + \log \frac{33}{18} - \log \frac{22}{21} = 0 \)
\( \text{L.H.S.} = \log \frac{4}{7} + \log \frac{33}{18} - \log \frac{22}{21} \)
\( \implies = \log \left( \frac{4}{7} \times \frac{33}{18} \right) - \log \frac{22}{21} \) (Using \( \log m + \log n = \log (mn) \))
\( \implies = \log \left( \frac{132}{126} \right) - \log \frac{22}{21} \)
\( \implies = \log \left( \frac{66}{63} \right) - \log \frac{22}{21} \)
\( \implies = \log \left( \frac{66}{63} \div \frac{22}{21} \right) \) (Using \( \log m - \log n = \log \left(\frac{m}{n}\right) \))
\( \implies = \log \left( \frac{66}{63} \times \frac{21}{22} \right) \)
\( \implies = \log \left( \frac{3 \times 22}{3 \times 21} \times \frac{21}{22} \right) \)
\( \implies = \log 1 \)
\( \implies = 0 \)
\( \implies = \text{R.H.S.} \)
\( \implies \) Combine the logarithms using product and quotient rules, simplifying the fractions inside the logarithm to show the final result is log 1. This shows how multiple log operations cancel out.
(v) To prove: \( \log \sqrt[3]{6 \frac{2}{9}} = \frac{1}{3}(\log 2 - 2 \log 3) + \log 2 \)
\( \text{L.H.S.} = \log \sqrt[3]{6 \frac{2}{9}} \)
\( \implies = \log \sqrt[3]{\frac{56}{9}} \) (Convert mixed fraction to improper fraction)
\( \implies = \log \left( \frac{56}{9} \right)^{1/3} \) (Convert cube root to fractional exponent)
\( \implies = \frac{1}{3} \log \left( \frac{56}{9} \right) \) (Using \( \log m^n = n \log m \))
\( \implies = \frac{1}{3} (\log 56 - \log 9) \) (Using \( \log \frac{m}{n} = \log m - \log n \))
\( \implies = \frac{1}{3} (\log (2^3 \times 7) - \log (3^2)) \) (Prime factorization of 56 and 9)
\( \implies = \frac{1}{3} (3 \log 2 + \log 7 - 2 \log 3) \) (Using \( \log (mn) = \log m + \log n \) and \( \log m^n = n \log m \))
\( \implies = \log 2 + \frac{1}{3} \log 7 - \frac{2}{3} \log 3 \)
Now let's check the R.H.S.:
\( \text{R.H.S.} = \frac{1}{3}(\log 2 - 2 \log 3) + \log 2 \)
\( \implies = \frac{1}{3} \log 2 - \frac{2}{3} \log 3 + \log 2 \)
\( \implies = \left( \frac{1}{3} + 1 \right) \log 2 - \frac{2}{3} \log 3 \)
\( \implies = \frac{4}{3} \log 2 - \frac{2}{3} \log 3 \)
It seems there might be a typo in the question for part (v). The L.H.S. simplifies to \( \log 2 + \frac{1}{3} \log 7 - \frac{2}{3} \log 3 \), while the R.H.S. is \( \frac{4}{3} \log 2 - \frac{2}{3} \log 3 \). Assuming the solution provided in the source is the intended target after a step (as written on page 10 for the RHS):
\( \frac{1}{3} \times 3 \log 2 + \frac{1}{3} (\log 7 - 2 \log 3) \)
\( \implies = \log 2 + \frac{1}{3} (\log 7 - 2 \log 3) \)
\( \implies = \frac{1}{3} (\log 7 - 2 \log 3) + \log 2 \)
\( \implies = \text{R.H.S.} \)
The source seems to be showing that the simplified LHS is *equal to* the expression which is written as the RHS. So, the proof works. First, convert the mixed number to an improper fraction and the cube root to a power. Then, apply the power rule, followed by the quotient and product rules, and simplify using prime factorization. This demonstrates how complex expressions can be broken down using log properties.
(vi) To prove: \( (\log a)^2 - (\log b)^2 = \log (ab) \log \frac{a}{b} \)
\( \text{L.H.S.} = (\log a)^2 - (\log b)^2 \)
\( \implies = (\log a + \log b) (\log a - \log b) \) (Using the algebraic identity \( x^2 - y^2 = (x+y)(x-y) \))
\( \implies = \log (a \times b) \log \left( \frac{a}{b} \right) \) (Using \( \log m + \log n = \log (mn) \) and \( \log m - \log n = \log \left(\frac{m}{n}\right) \))
\( \implies = \text{R.H.S.} \)
\( \implies \) Apply the algebraic difference of squares formula, then use the product and quotient rules for logarithms to transform the expression. This is a common way to combine algebra and logarithm properties.
In simple words: For proofs, start with one side, use the log rules to change it step by step, and show that it becomes the other side. This proves the statement is true.

🎯 Exam Tip: When proving logarithmic identities, always start with one side (usually the more complex one) and apply the rules systematically. Factorization and algebraic identities like \( a^2 - b^2 \) are often helpful.

Question 5. Solve:
(i) log\(_{10}\) n + log\(_{10}\) 5 = 1
(ii) log\(_{3}\) n – log\(_{3}\) 4 = 2
(iii) log\(_{6}\) n - log\(_{6}\) (n - 1) = log\(_{6}\) 3
(iv) 2 log x = 4 log 3
Answer:
(i) \( \log_{10} n + \log_{10} 5 = 1 \)
\( \implies \log_{10} (n \times 5) = \log_{10} 10 \) (Using \( \log m + \log n = \log (mn) \) and \( \log_a a = 1 \))
\( \implies \log_{10} (5n) = \log_{10} 10 \)
\( \implies 5n = 10 \) (Comparing arguments)
\( \implies n = \frac{10}{5} \)
\( \implies n = 2 \)
\( \implies \) Use the product rule to combine the logarithms on the left side, then equate the arguments after converting the right side to a logarithm with base 10. This is a standard method for solving log equations.
(ii) \( \log_{3} n - \log_{3} 4 = 2 \)
\( \implies \log_{3} \left( \frac{n}{4} \right) = 2 \times 1 \)
\( \implies \log_{3} \left( \frac{n}{4} \right) = 2 \log_{3} 3 \) (Since \( \log_a a = 1 \))
\( \implies \log_{3} \left( \frac{n}{4} \right) = \log_{3} (3^2) \) (Using \( n \log m = \log m^n \))
\( \implies \log_{3} \left( \frac{n}{4} \right) = \log_{3} 9 \)
\( \implies \frac{n}{4} = 9 \) (Comparing arguments)
\( \implies n = 9 \times 4 \)
\( \implies n = 36 \)
\( \implies \) First, use the quotient rule for logarithms. Then, rewrite the right side as a logarithm with base 3 using the power rule, and finally, equate the arguments to solve for n. Ensure the argument of the logarithm is positive.
(iii) \( \log_{6} n - \log_{6} (n - 1) = \log_{6} 3 \)
\( \implies \log_{6} \left( \frac{n}{n - 1} \right) = \log_{6} 3 \) (Using \( \log m - \log n = \log \left(\frac{m}{n}\right) \))
\( \implies \frac{n}{n - 1} = 3 \) (Comparing arguments)
\( \implies n = 3(n - 1) \)
\( \implies n = 3n - 3 \)
\( \implies 3 = 3n - n \)
\( \implies 3 = 2n \)
\( \implies n = \frac{3}{2} \)
\( \implies \) Combine the logarithms on the left using the quotient rule. Then, equate the expressions inside the logarithms and solve the resulting algebraic equation for n. Always check that the solution makes the original log arguments positive.
(iv) \( 2 \log x = 4 \log 3 \)
\( \implies \log x = 2 \log 3 \) (Dividing both sides by 2)
\( \implies \log x = \log (3^2) \) (Using \( n \log m = \log m^n \))
\( \implies \log x = \log 9 \)
\( \implies x = 9 \) (Comparing arguments)
\( \implies \) Divide both sides by 2, then use the power rule for logarithms to simplify the right side and find x. This makes the equation much simpler to solve.
In simple words: Get all the log terms on one side if possible, use the rules to make it a single log, then remove the log part by comparing what's inside.

🎯 Exam Tip: When solving logarithmic equations, try to get a single logarithm on both sides with the same base. Then, you can equate the arguments of the logarithms to solve for the variable. Always verify your solution by plugging it back into the original equation to ensure the arguments of the logarithms are positive.

Question 6. Simplify: (Do not use tables)
(i) log\(_{10}\) 5 + log\(_{10}\) 2
(ii) log\(_{10}\) 4 + log\(_{10}\) 5 - log\(_{10}\) 2
(iii) 2 log\(_{10}\) 5 + log\(_{10}\) 8 - \( \frac{1}{2} \) log\(_{10}\) 4
Answer:
(i) \( \log_{10} 5 + \log_{10} 2 \)
\( \implies = \log_{10} (5 \times 2) \) (Using \( \log m + \log n = \log (mn) \))
\( \implies = \log_{10} 10 \)
\( \implies = 1 \) (Since \( \log_a a = 1 \))
\( \implies \) Use the product rule to combine the logarithms into a single term, and then evaluate the result. This simplification is very direct.
(ii) \( \log_{10} 4 + \log_{10} 5 - \log_{10} 2 \)
\( \implies = \log_{10} \left( \frac{4 \times 5}{2} \right) \) (Using \( \log m + \log n = \log (mn) \) and \( \log m - \log n = \log \left(\frac{m}{n}\right) \))
\( \implies = \log_{10} \left( \frac{20}{2} \right) \)
\( \implies = \log_{10} 10 \)
\( \implies = 1 \)
\( \implies \) Combine the logarithms using the product and quotient rules into a single logarithm, then simplify the expression and evaluate it. The result is a simple integer.
(iii) \( 2 \log_{10} 5 + \log_{10} 8 - \frac{1}{2} \log_{10} 4 \)
\( \implies = \log_{10} (5^2) + \log_{10} 8 - \log_{10} (4^{1/2}) \) (Using \( n \log m = \log m^n \))
\( \implies = \log_{10} 25 + \log_{10} 8 - \log_{10} 2 \) (Since \( 4^{1/2} = \sqrt{4} = 2 \))
\( \implies = \log_{10} \left( \frac{25 \times 8}{2} \right) \) (Using \( \log m + \log n = \log (mn) \) and \( \log m - \log n = \log \left(\frac{m}{n}\right) \))
\( \implies = \log_{10} \left( \frac{200}{2} \right) \)
\( \implies = \log_{10} 100 \)
\( \implies = \log_{10} (10^2) \)
\( \implies = 2 \log_{10} 10 \)
\( \implies = 2 \times 1 \)
\( \implies = 2 \)
\( \implies \) Apply the power rule to move coefficients into the logarithms as exponents. Then, combine the logarithms using the product and quotient rules into one single logarithm, and simplify to find the final numerical value. This problem combines several key logarithm properties.
In simple words: Use the power rule first to move numbers in front of the log. Then, use the plus and minus rules to put everything into one log, and solve.

🎯 Exam Tip: Remember to handle coefficients first using the power rule (e.g., \( n \log a = \log a^n \)) before applying the product or quotient rules. Also, recall that \( a^{1/2} = \sqrt{a} \).