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Logarithms - Introduction to Logarithms-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of Logarithm: A logarithm is the power to which a base must be raised to yield a given number. If ax=ya^x = y, then log⁡ay=x\log_a y = x, where a>0a > 0, a≠1a \neq 1, and y>0y > 0.

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Fundamental Identities: The logarithm of 11 to any base is 00 (log⁡a1=0\log_a 1 = 0) and the logarithm of a number to the same base is 11 (log⁡aa=1\log_a a = 1).

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Product and Quotient Rules: The log of a product is the sum of the logs (log⁡a(mn)=log⁡am+log⁡an\log_a (mn) = \log_a m + \log_a n), and the log of a quotient is the difference (log⁡a(mn)=log⁡am−log⁡an\log_a (\frac{m}{n}) = \log_a m - \log_a n).

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Power Rule: The log of a number raised to an exponent is the exponent multiplied by the log of the number (log⁡amn=nlog⁡am\log_a m^n = n \log_a m).

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Change of Base Formula: This allows converting a logarithm to a different base: log⁡ba=log⁡calog⁡cb\log_b a = \frac{\log_c a}{\log_c b}. A common variation is log⁡ba=1log⁡ab\log_b a = \frac{1}{\log_a b}.

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Base Constraints: For log⁡ax\log_a x to be defined, the base aa must be positive and not equal to 11, and the argument xx must be strictly positive.

📐Formulae

log⁡a(m×n)=log⁡am+log⁡an\log_a (m \times n) = \log_a m + \log_a n

log⁡a(mn)=log⁡am−log⁡an\log_a \left(\frac{m}{n}\right) = \log_a m - \log_a n

log⁡a(mn)=nlog⁡am\log_a (m^n) = n \log_a m

log⁡ab=log⁡xblog⁡xa\log_a b = \frac{\log_x b}{\log_x a}

alog⁡ax=xa^{\log_a x} = x

log⁡akm=1klog⁡am\log_{a^k} m = \frac{1}{k} \log_a m

💡Examples

Problem 1:

Solve for xx: log⁡2(x−3)=4\log_2 (x - 3) = 4

Solution:

  1. Convert the logarithmic equation to exponential form using the definition ay=xa^y = x.
  2. 24=x−32^4 = x - 3
  3. 16=x−316 = x - 3
  4. x=16+3=19x = 16 + 3 = 19

Explanation:

To solve a basic logarithmic equation, we rewrite it in its exponential form where the base of the log becomes the base of the power.

Problem 2:

Simplify the expression: 2log⁡5+log⁡8−12log⁡42\log 5 + \log 8 - \frac{1}{2}\log 4

Solution:

  1. Apply the Power Rule: log⁡52+log⁡8−log⁡41/2\log 5^2 + \log 8 - \log 4^{1/2}
  2. Simplify powers: log⁡25+log⁡8−log⁡2\log 25 + \log 8 - \log 2
  3. Apply the Product Rule: log⁡(25×8)−log⁡2=log⁡200−log⁡2\log (25 \times 8) - \log 2 = \log 200 - \log 2
  4. Apply the Quotient Rule: log⁡(2002)=log⁡100\log (\frac{200}{2}) = \log 100
  5. Since the base is 1010 (common log), log⁡10102=2\log_{10} 10^2 = 2

Explanation:

The laws of logarithms are used sequentially to condense the expression into a single term before evaluating.

Problem 3:

If log⁡ax=p\log_a x = p and log⁡bx=q\log_b x = q, find the value of log⁡abx\log_{ab} x in terms of pp and qq.

Solution:

  1. Express the given equations as: 1p=log⁡xa\frac{1}{p} = \log_x a and 1q=log⁡xb\frac{1}{q} = \log_x b.
  2. We need log⁡abx\log_{ab} x, which can be written as 1log⁡x(ab)\frac{1}{\log_x (ab)}.
  3. Using the Product Rule: log⁡x(ab)=log⁡xa+log⁡xb\log_x (ab) = \log_x a + \log_x b.
  4. Substitute the values: log⁡x(ab)=1p+1q=p+qpq\log_x (ab) = \frac{1}{p} + \frac{1}{q} = \frac{p + q}{pq}.
  5. Therefore, log⁡abx=1p+qpq=pqp+q\log_{ab} x = \frac{1}{\frac{p+q}{pq}} = \frac{pq}{p+q}.

Explanation:

This problem uses the reciprocal property of the change of base formula to handle different bases sharing the same argument.