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Logarithms - Understanding Logarithms as the Inverse of Exponents-advanced

Grade 9CBSE

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

🔑Concepts

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Definition: The logarithm of a number NN to a base aa is the exponent xx to which the base aa must be raised to obtain NN. This is written as log⁡aN=x\log_a N = x and is equivalent to ax=Na^x = N.

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Constraints: For the expression log⁡aN\log_a N to be defined, the base aa must be positive and not equal to 11 (a>0,a≠1a > 0, a \neq 1), and the argument NN must be positive (N>0N > 0).

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Inverse Relationship: Logarithmic functions are the inverse of exponential functions. This leads to the identity alog⁡ax=xa^{\log_a x} = x and log⁡a(ax)=x\log_a(a^x) = x.

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Common Logarithms: Logarithms with base 1010 are known as common logarithms. When the base is not written, it is often assumed to be 1010 (log⁡N=log⁡10N\log N = \log_{10} N).

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Natural Logarithms: Logarithms with the base ee (Euler's number ≈2.718\approx 2.718) are called natural logarithms, denoted as ln⁡x\ln x or log⁡ex\log_e x.

📐Formulae

log⁡aN=x  ⟺  ax=N\log_a N = x \iff a^x = N

log⁡a(M⋅N)=log⁡aM+log⁡aN\log_a (M \cdot 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(Mk)=klog⁡aM\log_a (M^k) = k \log_a M

log⁡aa=1\log_a a = 1

log⁡a1=0\log_a 1 = 0

log⁡ba=log⁡calog⁡cb (Change of Base formula)\log_b a = \frac{\log_c a}{\log_c b} \text{ (Change of Base formula)}

log⁡ab=1log⁡ba\log_a b = \frac{1}{\log_b a}

💡Examples

Problem 1:

Evaluate the value of xx if log⁡381=x\log_{\sqrt{3}} 81 = x.

Solution:

  1. Convert the logarithmic equation to its exponential form: (3)x=81(\sqrt{3})^x = 81
  2. Express both sides with the same base (base 33): (31/2)x=34(3^{1/2})^x = 3^4
  3. Use the power of a power rule (am)n=amn(a^m)^n = a^{mn}: 3x/2=343^{x/2} = 3^4
  4. Since the bases are the same, equate the exponents: x2=4\frac{x}{2} = 4 x=8x = 8

Explanation:

By converting the radical base into a power of 33 and writing 8181 as 343^4, we can solve for the exponent directly.

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 klog⁡M=log⁡Mkk \log M = \log M^k to each term: log⁡52+log⁡8−log⁡41/2\log 5^2 + \log 8 - \log 4^{1/2} log⁡25+log⁡8−log⁡2\log 25 + \log 8 - \log 2
  2. Apply the Product Rule log⁡M+log⁡N=log⁡(MN)\log M + \log N = \log(MN): log⁡(25×8)−log⁡2\log(25 \times 8) - \log 2 log⁡200−log⁡2\log 200 - \log 2
  3. Apply the Quotient Rule log⁡M−log⁡N=log⁡(MN)\log M - \log N = \log(\frac{M}{N}): log⁡(2002)\log \left(\frac{200}{2}\right) log⁡100\log 100
  4. Evaluate the common logarithm (base 1010): log⁡10102=2\log_{10} 10^2 = 2

Explanation:

We use the fundamental laws of logarithms to condense the multiple terms into a single logarithmic value, then simplify using the base 1010 property.

Problem 3:

Solve for xx: log⁡x(164)=−3\log_x \left(\frac{1}{64}\right) = -3.

Solution:

  1. Convert the logarithmic equation to exponential form: x−3=164x^{-3} = \frac{1}{64}
  2. Write 164\frac{1}{64} as a power with a negative exponent: x−3=64−1x^{-3} = 64^{-1}
  3. Since 64=4364 = 4^3, we can write: x−3=(43)−1=4−3x^{-3} = (4^3)^{-1} = 4^{-3}
  4. Comparing both sides: x=4x = 4

Explanation:

The exponential form x−3=164x^{-3} = \frac{1}{64} allows us to equate the bases by matching the exponents.