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Algebra - Logarithms

Grade 9ICSE

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

🔑Concepts

Definition of Logarithm: A logarithm is the inverse operation of exponentiation. If ax=ya^x = y, then we say that the logarithm of yy to the base aa is xx, written as logay=x\log_a y = x. This relationship connects the base aa, the exponent xx, and the resulting value yy. Visually, if you imagine an exponential growth curve rising steeply, the logarithm reflects this across the line y=xy=x, showing how the exponent grows slowly as the value increases.

Existence Conditions: For a logarithm logam\log_a m to be defined, the base aa must be a positive real number not equal to 11 (a>0,a1a > 0, a \neq 1), and the number mm must be positive (m>0m > 0). On a coordinate plane, the graph of y=logaxy = \log_a x only exists to the right of the y-axis, indicating that we cannot take the logarithm of zero or negative numbers.

Fundamental Identities: There are two critical values to remember. First, logaa=1\log_a a = 1 because any number raised to the power of 11 is itself (a1=aa^1 = a). Second, loga1=0\log_a 1 = 0 because any non-zero number raised to the power of 00 is 11 (a0=1a^0 = 1). This means the graph of every logarithmic function y=logaxy = \log_a x passes through the point (1,0)(1, 0) on the x-axis.

Product and Quotient Rules: Logarithms convert multiplication into addition and division into subtraction. Specifically, the log of a product is the sum of the logs: loga(mn)=logam+logan\log_a (mn) = \log_a m + \log_a n. Conversely, the log of a quotient is the difference of the logs: loga(mn)=logamlogan\log_a (\frac{m}{n}) = \log_a m - \log_a n. This effectively 'compresses' operations, making large-scale calculations easier to handle.

Power Rule: If the argument of a logarithm has an exponent, that exponent can be moved to the front of the logarithm as a multiplier: loga(mn)=nlogam\log_a (m^n) = n \log_a m. Visually, this shows that scaling the exponent of the input results in a linear scaling of the output value.

Common Logarithms: Logarithms with base 1010 are known as common logarithms. In many textbooks and calculators, if no base is specified (e.g., logx\log x), the base is assumed to be 1010. These are particularly useful for expressing numbers in scientific notation and calculating magnitudes.

Base Change and Reciprocal Property: The value of a logarithm can be expressed using a different base using the formula logam=logbmlogba\log_a m = \frac{\log_b m}{\log_b a}. Additionally, swapping the base and the argument results in a reciprocal: logab=1logba\log_a b = \frac{1}{\log_b a}.

📐Formulae

logay=x    ax=y\log_a y = x \iff a^x = y

loga(m×n)=logam+logan\log_a (m \times n) = \log_a m + \log_a n

loga(mn)=logamlogan\log_a (\frac{m}{n}) = \log_a m - \log_a n

loga(mn)=nlogam\log_a (m^n) = n \log_a m

logaa=1\log_a a = 1

loga1=0\log_a 1 = 0

alogax=xa^{\log_a x} = x

logam=logbmlogba\log_a m = \frac{\log_b m}{\log_b a}

💡Examples

Problem 1:

Evaluate the value of xx if log3(x2)=2\log_3 (x - 2) = 2.

Solution:

  1. Convert the logarithmic equation into its equivalent exponential form: 32=x23^2 = x - 2.
  2. Calculate the power: 9=x29 = x - 2.
  3. Solve for xx by adding 22 to both sides: x=9+2x = 9 + 2.
  4. Therefore, x=11x = 11.

Explanation:

To solve an equation where the variable is inside a logarithm, we isolate the log term and convert it to exponential form using the definition logay=x    ax=y\log_a y = x \implies a^x = y.

Problem 2:

Simplify and evaluate: 2log5+log812log42 \log 5 + \log 8 - \frac{1}{2} \log 4 (Base 10).

Solution:

  1. Use the Power Rule to move coefficients inside: log52+log8log41/2\log 5^2 + \log 8 - \log 4^{1/2}.
  2. Simplify the powers: log25+log8log2\log 25 + \log 8 - \log 2 (since 41/2=4=24^{1/2} = \sqrt{4} = 2).
  3. Use the Product Rule for addition: log(25×8)log2=log200log2\log (25 \times 8) - \log 2 = \log 200 - \log 2.
  4. Use the Quotient Rule for subtraction: log(2002)=log100\log (\frac{200}{2}) = \log 100.
  5. Since the base is 1010, evaluate log10100\log_{10} 100: log10102=2log1010=2(1)=2\log_{10} 10^2 = 2 \log_{10} 10 = 2(1) = 2.

Explanation:

This problem uses all the fundamental laws of logarithms (Power, Product, and Quotient) to condense multiple terms into a single logarithmic expression that can be easily evaluated.