OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Chapter Test

Question 1. The value of \( \log_2 16 \) is
(a) \( \frac { 1 }{8} \)
(b) 4
(c) 8
(d) 16
Answer: (b) 4
\( \log_2 16 \)
We need to find the power to which 2 must be raised to get 16.
\( \log_2 16 = \log_2 (2^4) \)
Using the logarithm property \( \log_b (b^x) = x \):
\( \log_2 (2^4) = 4 \)
Therefore, the value is 4. Logarithms help us find the exponent in an exponential equation.
In simple words: This question asks "2 to what power equals 16?". Since 2 multiplied by itself 4 times (2 x 2 x 2 x 2) makes 16, the answer is 4.

🎯 Exam Tip: Remember the basic definition of a logarithm: \( \log_b a = x \) means \( b^x = a \). Practice converting between these two forms quickly.

Question 2. If \( a^x = b^y \) then
(a) \( \log\frac{a}{b}=\frac{x}{y} \)
(b) \( \frac{\log a}{\log b}=\frac{x}{y} \)
(c) \( \frac{\log a}{\log b}=\frac{y}{x} \)
(d) None of these
Answer: (c) \( \frac{\log a}{\log b}=\frac{y}{x} \)
Given the equation: \( a^x = b^y \)
Take the logarithm on both sides of the equation:
\( \log (a^x) = \log (b^y) \)
Using the logarithm property \( \log (M^p) = p \log M \):
\( x \log a = y \log b \)
Now, rearrange the terms to form a ratio:
\( \frac{\log a}{\log b} = \frac{y}{x} \)
This property is very useful for solving equations with variables in the exponent.
In simple words: When two exponential terms are equal, taking log on both sides helps to bring down their powers. Then, you can make a fraction with the logs of the bases and the powers, like the answer shows.

🎯 Exam Tip: Always remember the power rule of logarithms, \( \log (M^p) = p \log M \), as it's key for solving equations where variables are exponents.

Question 3. If \( \log 3 = 0.477 \) and \( (1000)^x = 3 \), then \( x \) equals
(a) 0.0159
(b) 0.0477
(c) 0.159
(d) 10
Answer: (c) 0.159
Given: \( \log 3 = 0.477 \)
Given: \( (1000)^x = 3 \)
To find \( x \), take the logarithm on both sides of the second equation:
\( \log (1000)^x = \log 3 \)
Using the power rule of logarithms, \( \log (M^p) = p \log M \):
\( x \log 1000 = \log 3 \)
We know that \( \log 1000 = \log (10^3) = 3 \) (assuming base 10 for log).
So, substitute the known values into the equation:
\( x \times 3 = 0.477 \)
To solve for \( x \), divide 0.477 by 3:
\( x = \frac{0.477}{3} \)
\( x = 0.159 \)
This process converts an exponential problem into a simpler algebraic one.
In simple words: We are given the log of 3 and an equation with \( x \). First, we apply log to both sides of the \( 1000^x \) equation. Since log of 1000 is 3, we get \( 3x = \log 3 \). Then, we just divide \( \log 3 \) by 3 to find \( x \).

🎯 Exam Tip: Remember that \( \log_{10} 1000 = 3 \). Always look for ways to simplify numbers inside logarithms to base 10 powers if possible.

Question 4. If \( \log_{10} 2 = 0.3010 \), the value of \( \log_{10} 5 \) is
(a) 0.3241
(b) 0.6911
(c) 0.6990
(d) 0.7525
Answer: (c) 0.6990
Given: \( \log_{10} 2 = 0.3010 \)
We need to find \( \log_{10} 5 \).
We can write 5 as \( \frac{10}{2} \).
So, \( \log_{10} 5 = \log_{10} (\frac{10}{2}) \)
Using the logarithm property \( \log_b (\frac{M}{N}) = \log_b M - \log_b N \):
\( \log_{10} 5 = \log_{10} 10 - \log_{10} 2 \)
We know that \( \log_{10} 10 = 1 \).
Substitute the values:
\( \log_{10} 5 = 1 - 0.3010 \)
\( \log_{10} 5 = 0.6990 \)
This is a common trick to find the log of 5 using the log of 2 and 10.
In simple words: To find \( \log_{10} 5 \), think of 5 as 10 divided by 2. Then, use the log rule for division, which turns it into subtraction. Since \( \log_{10} 10 \) is 1, subtract \( \log_{10} 2 \) (which is given) from 1.

🎯 Exam Tip: Remember the relationship \( \log_{10} 5 = 1 - \log_{10} 2 \). This is a very frequent shortcut in logarithm problems.

Question 5. If \( \log_{10} 2 = 0.3010 \), the value of \( \log_{10} 80 \) is
(a) 1.6020
(b) 1.9030
(c) 3.9030
(d) None of these
Answer: (b) 1.9030
Given: \( \log_{10} 2 = 0.3010 \)
We need to find \( \log_{10} 80 \).
First, express 80 in terms of its prime factors and powers of 10:
\( 80 = 8 \times 10 = 2^3 \times 10 \)
So, \( \log_{10} 80 = \log_{10} (10 \times 2^3) \)
Using the logarithm property \( \log_b (M \times N) = \log_b M + \log_b N \):
\( \log_{10} 80 = \log_{10} 10 + \log_{10} (2^3) \)
Using the power rule of logarithms, \( \log (M^p) = p \log M \):
\( \log_{10} 80 = \log_{10} 10 + 3 \log_{10} 2 \)
We know that \( \log_{10} 10 = 1 \).
Substitute the known values:
\( \log_{10} 80 = 1 + 3 \times 0.3010 \)
\( \log_{10} 80 = 1 + 0.9030 \)
\( \log_{10} 80 = 1.9030 \)
This shows how to use log properties to simplify calculations for larger numbers.
In simple words: To find \( \log_{10} 80 \), first write 80 as \( 10 \times 2^3 \). Then, use the log rules for multiplication and powers. This gives \( \log_{10} 10 + 3 \log_{10} 2 \). Put in the given value for \( \log_{10} 2 \) and simplify.

🎯 Exam Tip: When dealing with numbers like 80, 800, 8000, always try to factor out powers of 10 (like 10, 100, 1000) as \( \log_{10} 10^n = n \), which simplifies calculations greatly.

Question 6. If \( \log_{10} 7 = a \), then \( \log_{10} (\frac { 1 }{ 70 }) \) is equal to
(a) \( - (1 + a) \)
(c) \( \frac { a }{ 10 } \)
(d) \( \frac { 1 }{ 10a } \)
Answer: (a) \( - (1 + a) \)
Given: \( \log_{10} 7 = a \)
We need to find \( \log_{10} (\frac{1}{70}) \).
Using the logarithm property \( \log_b (\frac{M}{N}) = \log_b M - \log_b N \):
\( \log_{10} (\frac{1}{70}) = \log_{10} 1 - \log_{10} 70 \)
We know that \( \log_{10} 1 = 0 \).
So, the expression becomes \( 0 - \log_{10} 70 = - \log_{10} 70 \)
Now, express 70 as a product of 7 and 10:
\( 70 = 7 \times 10 \)
So, \( - \log_{10} 70 = - \log_{10} (7 \times 10) \)
Using the logarithm property \( \log_b (M \times N) = \log_b M + \log_b N \):
\( - \log_{10} (7 \times 10) = - (\log_{10} 7 + \log_{10} 10) \)
Substitute \( \log_{10} 7 = a \) and \( \log_{10} 10 = 1 \):
\( - (a + 1) \)
This demonstrates the power of logarithmic rules to simplify complex expressions.
In simple words: To find \( \log_{10} (\frac{1}{70}) \), first use the log rule for division, which makes it \( \log_{10} 1 - \log_{10} 70 \). Since \( \log_{10} 1 \) is 0, we are left with \( -\log_{10} 70 \). Now, break 70 into \( 7 \times 10 \), and use the log rule for multiplication. This changes it to \( -(\log_{10} 7 + \log_{10} 10) \). Finally, substitute \( a \) for \( \log_{10} 7 \) and 1 for \( \log_{10} 10 \).

🎯 Exam Tip: Be careful with negative signs when distributing them across terms in parentheses, especially after using the division rule for logarithms.

Question 7. If \( \log 27 = 1.431 \), then the value of \( \log 9 \) is
(a) 0.934
(b) 0.945
(c) 0.954
(d) 0.958
Answer: (c) 0.954
Given: \( \log 27 = 1.431 \)
We can write 27 as \( 3^3 \).
So, \( \log (3^3) = 1.431 \)
Using the power rule of logarithms, \( \log (M^p) = p \log M \):
\( 3 \log 3 = 1.431 \)
Now, find the value of \( \log 3 \):
\( \log 3 = \frac{1.431}{3} \)
\( \log 3 = 0.477 \)
Next, we need to find \( \log 9 \). We can write 9 as \( 3^2 \).
So, \( \log 9 = \log (3^2) \)
Using the power rule again:
\( \log 9 = 2 \log 3 \)
Substitute the value of \( \log 3 \):
\( \log 9 = 2 \times 0.477 \)
\( \log 9 = 0.954 \)
This problem uses the exponent rule of logarithms in reverse.
In simple words: First, use the given \( \log 27 \) to find \( \log 3 \) because \( 27 \) is \( 3^3 \). So, \( 3 \log 3 = 1.431 \), which means \( \log 3 = 0.477 \). Then, use \( \log 9 \) which is \( \log (3^2) \), meaning \( 2 \log 3 \). Multiply 2 by 0.477 to get the answer.

🎯 Exam Tip: Always look for relationships between the numbers given (like 27 and 9 both being powers of 3) to apply logarithm properties effectively.

Question 8. If \( \log_{10} 5 + \log_{10}(5x + 1) = \log_{10}(x + 5) + 1 \), then \( x \) is equal to
(a) 1
(b) 3
(c) 5
(d) 10
Answer: (b) 3
Given the equation: \( \log_{10} 5 + \log_{10}(5x + 1) = \log_{10}(x + 5) + 1 \)
First, apply the logarithm product rule \( \log M + \log N = \log (M \times N) \) to the left side:
\( \log_{10} [5(5x + 1)] = \log_{10}(x + 5) + 1 \)
\( \log_{10} (25x + 5) = \log_{10}(x + 5) + 1 \)
Next, express the constant 1 as a logarithm with base 10: \( 1 = \log_{10} 10 \).
\( \log_{10} (25x + 5) = \log_{10}(x + 5) + \log_{10} 10 \)
Apply the product rule to the right side:
\( \log_{10} (25x + 5) = \log_{10} [10(x + 5)] \)
\( \log_{10} (25x + 5) = \log_{10} (10x + 50) \)
Since both sides have \( \log_{10} \), we can equate the arguments:
\( 25x + 5 = 10x + 50 \)
Now, solve the linear equation for \( x \):
Subtract \( 10x \) from both sides:
\( 25x - 10x + 5 = 50 \)
\( 15x + 5 = 50 \)
Subtract 5 from both sides:
\( 15x = 50 - 5 \)
\( 15x = 45 \)
Divide by 15:
\( x = \frac{45}{15} \)
\( x = 3 \)
This problem shows how to solve logarithmic equations by simplifying them.
In simple words: Combine the log terms on both sides using log rules for addition. Remember that 1 can be written as \( \log_{10} 10 \). Once you have \( \log A = \log B \), you can just set \( A = B \) and solve for \( x \) like a normal algebra problem.

🎯 Exam Tip: Before solving, make sure all terms are written as logarithms with the same base. If there's a constant, convert it to a logarithm using the base of the other terms.

Question 9. If \( \log_x 4 = 0.4 \), then the value of \( x \) is
(a) 1
(b) 4
(c) 16
(d) 32
Answer: (d) 32
Given the logarithmic equation: \( \log_x 4 = 0.4 \)
Convert this logarithmic form to its equivalent exponential form. The base of the logarithm (\( x \)) becomes the base of the exponent, and the result of the logarithm (0.4) becomes the exponent:
\( x^{0.4} = 4 \)
To solve for \( x \), raise both sides of the equation to the reciprocal of the exponent, which is \( \frac{1}{0.4} \):
\( (x^{0.4})^{\frac{1}{0.4}} = 4^{\frac{1}{0.4}} \)
\( x = 4^{\frac{10}{4}} \)
\( x = 4^{2.5} \)
Now, calculate \( 4^{2.5} \). We can write 4 as \( 2^2 \):
\( x = (2^2)^{2.5} \)
Using the exponent rule \( (a^m)^n = a^{m \times n} \):
\( x = 2^{2 \times 2.5} \)
\( x = 2^5 \)
\( x = 32 \)
This shows how to convert between logarithmic and exponential forms to solve equations.
In simple words: Change the log equation \( \log_x 4 = 0.4 \) into an exponent equation: \( x^{0.4} = 4 \). To get \( x \) by itself, raise both sides to the power of \( \frac{1}{0.4} \) (which is 2.5). Then, calculate \( 4^{2.5} \), which equals 32.

🎯 Exam Tip: Always remember that \( a^{m/n} = \sqrt[n]{a^m} \). This helps in understanding and calculating fractional exponents like \( 4^{2.5} = 4^{5/2} = (\sqrt{4})^5 = 2^5 = 32 \).

Question 10. The solution of \( \log (\log_2 (\log_7 x)) = 0 \) is
(a) 2
(b) \( \pi^2 \)
(c) 72
(d) None of these
Answer: (d) None of these
Given the equation: \( \log (\log_2 (\log_7 x)) = 0 \)
Let the base of the outermost logarithm be \( B \). (For any base \( B \), if \( \log_B A = 0 \), then \( A = B^0 = 1 \)).
So, the argument of the outermost log must be 1:
\( \log_2 (\log_7 x) = 1 \)
Next, convert this logarithmic form to exponential form: \( \log_b A = x \implies b^x = A \).
The base is 2, and the result is 1, so:
\( \log_7 x = 2^1 \)
\( \log_7 x = 2 \)
Finally, convert this logarithmic form to exponential form:
The base is 7, and the result is 2, so:
\( x = 7^2 \)
\( x = 49 \)
Since 49 is not listed in the options, the correct answer is 'None of these'. This problem requires careful unwrapping of multiple logarithm layers.
In simple words: This problem has logs inside logs. Start from the outside: if log of something is 0, that 'something' must be 1. So, \( \log_2 (\log_7 x) \) is 1. Next, if \( \log_2 \) of something is 1, that 'something' must be \( 2^1 \), which is 2. So, \( \log_7 x = 2 \). Finally, if \( \log_7 x = 2 \), then \( x \) must be \( 7^2 \), which is 49.

🎯 Exam Tip: When solving nested logarithm equations, always work from the outermost logarithm inwards, converting each layer into its exponential form step-by-step. Be careful not to skip any steps.