Logarithm of a Power

logb ⁣(xn)=nlogbx\log_b\!\left(x^{n}\right) = n \log_b x

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Repeated multiplication becomes repeated addition under a logarithm, so log_b(xⁿ) is just n copies of log_b x. Worked example: log₃(9⁵) = 5 log₃9 = 5 × 2 = 10, and indeed 9⁵ = 59049 = 3¹⁰. Running it backwards solves for the base of a power: if log₂(x⁴) = 12 then x⁴ = 2¹² = 4096, so x = 8.

This is the identity that makes logarithms a method rather than a curiosity, because it is the only tool that brings an unknown down out of an exponent. Faced with 3ˣ = 50, take logs of both sides: x log 3 = log 50, so x = log 50 / log 3 ≈ 3.56. Every compound-interest "how many years?" question, every half-life calculation, and every algorithm analysis that argues a binary search takes log₂n steps rests on this one move. The trap is bracketing: log(xⁿ) means the whole power is inside the log, whereas (log x)ⁿ raises the logarithm itself — log₂(8²) = 6, but (log₂8)² = 9. And x must be positive, so log(x²) = 2 log x holds only for x > 0, even though x² is happy with negatives.

Logarithm of a Power
logb ⁣(xn)=nlogbx\log_b\!\left(x^{n}\right) = n \log_b x
Where
  • LL= Log of the power
  • xx= Base of the power
  • nn= Exponent
  • bb= Logarithm base
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