Logarithms

What is a logarithm?

A logarithm asks what power a base must be raised to in order to produce a number: 24=16  ⟺  log⁡216=42^4 = 16 \iff \log_2 16 = 4. Likewise 64=12966^4 = 1296, so log⁡61296=4\log_6 1296 = 4.

The logarithm of a number between 0 and 1 is negative (log⁡100.01=−2\log_{10} 0.01 = -2), and the logarithm of zero or a negative number is undefined for real numbers.

Common, natural and binary logarithms

  • Common (base 10): log⁡101000=3\log_{10} 1000 = 3.
  • Natural (base e≈2.71828e \approx 2.71828), written ln⁡x\ln x: it appears in continuous growth, such as FV=PV⋅eitFV = PV \cdot e^{it}.
  • Binary (base 2), common in computer science: log⁡28=3\log_2 8 = 3.

Logarithm rules

log⁡b(xy)=log⁡bx+log⁡by,log⁡b ⁣(xy)=log⁡bx−log⁡by,log⁡b(xy)=ylog⁡bx\log_b(xy) = \log_b x + \log_b y, \qquad \log_b\!\left(\dfrac{x}{y}\right) = \log_b x - \log_b y, \qquad \log_b(x^y) = y\log_b xlog⁡bx=log⁡kxlog⁡kb    (change of base),blog⁡bx=x\log_b x = \dfrac{\log_k x}{\log_k b} \;\;\text{(change of base)}, \qquad b^{\log_b x} = x

For example, log⁡832=log⁡232log⁡28=53\log_8 32 = \dfrac{\log_2 32}{\log_2 8} = \dfrac{5}{3}.

Worked example: solving a log equation

Solve log⁡2x+log⁡2(x−2)=3\log_2 x + \log_2(x-2) = 3.

Solution: Combine the logs: log⁡2[x(x−2)]=3\log_2[x(x-2)] = 3, so x(x−2)=8x(x-2) = 8, that is x2−2x−8=0x^2 - 2x - 8 = 0, giving x=4x = 4 or x=−2x = -2. Only x=4x = 4 works: x=−2x = -2 would need log⁡2(−2)\log_2(-2), which is undefined. Check: log⁡24+log⁡22=2+1=3\log_2 4 + \log_2 2 = 2 + 1 = 3.

Worked example: counting digits

How many digits does 2502^{50} have, given log⁡102≈0.30103\log_{10} 2 \approx 0.30103?

Solution: log⁡10250=50×0.30103=15.05\log_{10} 2^{50} = 50 \times 0.30103 = 15.05. A number NN with log⁡10N\log_{10} N between 15 and 16 has 16 digits. (Indeed 250=1,125,899,906,842,6242^{50} = 1{,}125{,}899{,}906{,}842{,}624.)

Worked example: simplifying with the rules

Each of these collapses to a whole number once the rules are applied in the right order:

log⁡102+log⁡105=log⁡1010=1,log⁡381−log⁡39=log⁡39=2\log_{10} 2 + \log_{10} 5 = \log_{10} 10 = 1, \qquad \log_3 81 - \log_3 9 = \log_3 9 = 22log⁡63+log⁡64−log⁡66=log⁡69×46=log⁡66=12\log_6 3 + \log_6 4 - \log_6 6 = \log_6\dfrac{9 \times 4}{6} = \log_6 6 = 1

Worked example: an exponential equation

Solve 3x=203^x = 20. Take logarithms of both sides and bring the power down: xlog⁡3=log⁡20x\log 3 = \log 20, so x=log⁡20log⁡3≈2.727x = \dfrac{\log 20}{\log 3} \approx 2.727. This is the change-of-base formula in action: x=log⁡320x = \log_3 20.

Common mistakes

  • Splitting a sum. log⁡(x+y)\log(x + y) is not log⁡x+log⁡y\log x + \log y; the rule is for a product.
  • Keeping invalid roots. Check every solution in the original equation's domain.
  • Inverting change of base. It's log⁡xlog⁡b\dfrac{\log x}{\log b}, with the base underneath.
  • Assuming log⁡\log means ln⁡\ln. Check which base the question uses.

Where this comes up in exams

Powers, exponents and logarithms are listed under numerical computation in the General Aptitude syllabus of every GATE 2027 paper. JEE Advanced 2026 lists logarithms and their properties, and JEE Main 2026 lists logarithmic functions in its limits, continuity and differentiability unit.

Checked against GATE 2027 information brochure (General Aptitude syllabus), JEE (Advanced) 2026 syllabus and JEE (Main) 2026 syllabus (NTA). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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Logarithms undo exponents, and they also solve growth and decay differential equations.

From past GATE papers