Introduction

A power ana^n is repeated multiplication; an exponent is the number that says how many times; a logarithm is the exponent written as the answer. These three views of one relation, ax=y  ⟺  x=log⁡aya^x = y \iff x = \log_a y, are the tools for the standard numerical tasks: simplify an expression with mixed bases, solve an equation where the unknown sits in the exponent, compare two huge numbers, find a units digit, count the digits of a power, or find how many periods of compound growth reach a target.

The methods are few and mechanical. The traps are in the rules that look true but are not, such as ln⁡(a+b)=ln⁡a+ln⁡b\ln(a + b) = \ln a + \ln b or (ab)c=a(bc)(a^b)^c = a^{(b^c)}, and in rounding a logarithm too early.

Prerequisites

  • Integer arithmetic, prime factorisation, and fractions.
  • For growth questions, the multiplier view of percentage change from Ratios and Percentages.

Core concepts

TermDefinitionExample
Poweran=a×a×⋯×aa^n = a \times a \times \cdots \times a (nn factors) for a positive integer nn25=322^5 = 32
Zero exponenta0=1a^0 = 1 for a≠0a \ne 070=17^0 = 1
Negative exponenta−n=1/ana^{-n} = 1 / a^n for a≠0a \ne 02−3=1/82^{-3} = 1/8
Fractional exponentap/q=(aq)pa^{p/q} = (\sqrt[q]{a})^p for a>0a > 0272/3=927^{2/3} = 9
Logarithmlog⁡ay=x\log_a y = x means ax=ya^x = y, for a>0a > 0, a≠1a \ne 1, y>0y > 0log⁡232=5\log_2 32 = 5
Common and natural logslog⁡10\log_{10} and ln⁡=log⁡e\ln = \log_e; these notes write lg⁡\lg for log⁡2\log_2log⁡101000=3\log_{10} 1000 = 3
SurdAn irrational root such as 2\sqrt{2} or 53\sqrt[3]{5}50=52\sqrt{50} = 5\sqrt{2}

Domain convention. Fractional exponents and logarithms are used only with positive bases and positive arguments in these notes. 000^0 is left undefined, and (−8)1/3(-8)^{1/3} is avoided because the rule ap/q=(a1/q)pa^{p/q} = (a^{1/q})^p breaks for negative aa with some representations of the same fraction.

The laws

For a,b>0a, b > 0 and real m,nm, n:

LawStatement
Productaman=am+na^m a^n = a^{m+n}
Quotientam/an=am−na^m / a^n = a^{m-n}
Power of a power(am)n=amn(a^m)^n = a^{mn}
Power of a product(ab)n=anbn(ab)^n = a^n b^n
Log of a productlog⁡c(xy)=log⁡cx+log⁡cy\log_c (xy) = \log_c x + \log_c y
Log of a quotientlog⁡c(x/y)=log⁡cx−log⁡cy\log_c (x/y) = \log_c x - \log_c y
Log of a powerlog⁡c(xk)=klog⁡cx\log_c (x^k) = k \log_c x
Change of baselog⁡ay=log⁡cy/log⁡ca\log_a y = \log_c y / \log_c a for any valid base cc
Inverse pairalog⁡ay=ya^{\log_a y} = y and log⁡a(ax)=x\log_a (a^x) = x
Reciprocallog⁡ab=1/log⁡ba\log_a b = 1 / \log_b a

Each log law is an exponent law read backwards: if x=cux = c^u and y=cvy = c^v, then xy=cu+vxy = c^{u+v}, so log⁡c(xy)=u+v\log_c(xy) = u + v.

Decision procedure

  1. Same base first. Rewrite every number as a power of a common prime: 4=224 = 2^2, 8=238 = 2^3, 27=3327 = 3^3, 125=53125 = 5^3. Then exponents simply add and subtract.
  2. Unknown in the exponent. Bring both sides to the same base and equate exponents; if that is impossible, take logarithms of both sides.
  3. Comparing two powers. Make the exponents equal (2300=81002^{300} = 8^{100}, 3200=91003^{200} = 9^{100}) or the bases equal; failing both, compare nln⁡an \ln a.
  4. Units digit. For positive integer exponents, units digits of powers repeat with period 1, 2 or 4. Reduce the exponent modulo 4 (use 4 when the remainder is 0).
  5. Number of digits. A positive integer NN has ⌊log⁡10N⌋+1\lfloor \log_{10} N \rfloor + 1 digits.
  6. Growth to a target. For m>1m>1, k≥1k\ge1 and a nonnegative integer number of periods nn, the least nn with mn≥km^n\ge k is ⌈ln⁡k/ln⁡m⌉\lceil\ln k/\ln m\rceil. Confirm with the powers on either side. If k≤1k\le1, the target is already reached at n=0n=0. For m=1m=1 or decay 0<m<10<m<1, handle reachability separately; do not apply this ceiling blindly.
  7. Logarithmic equations. Combine logs into one, convert to exponential form, solve, then reject any root that makes an argument non-positive.

Mathematical explanation

Why units digits cycle. Handle exponent 0 first: a0=1a^0=1 for a≠0a\ne0; 000^0 is undefined under this note's convention. For positive integer exponents: The units digit of ana^n depends only on the units digit of aa and is obtained by multiplying the previous units digit by it. There are at most ten digits, so the sequence repeats; checking the ten cases shows every cycle length divides 4:

Last digit of aaUnits digits of a1,a2,a3,a4a^1, a^2, a^3, a^4Period
0, 1, 5, 6the digit itself1
44, 6, 4, 62
99, 1, 9, 12
22, 4, 8, 64
33, 9, 7, 14
77, 9, 3, 14
88, 4, 2, 64

Digit count. If NN has dd digits then 10d−1≤N<10d10^{d-1} \le N < 10^d, so d−1≤log⁡10N<dd - 1 \le \log_{10} N < d, which gives d=⌊log⁡10N⌋+1d = \lfloor \log_{10} N \rfloor + 1.

Change of base. Let x=log⁡ayx = \log_a y, so ax=ya^x = y. Taking log⁡c\log_c of both sides, xlog⁡ca=log⁡cyx \log_c a = \log_c y, which is the change-of-base rule.

Growth. For m>1m>1, k≥1k\ge1, solving mn=km^n=k gives n=ln⁡k/ln⁡mn=\ln k/\ln m; the first nonnegative whole period at or above the target is its ceiling. For k≤1k\le1, use zero periods; constant or decaying multipliers need a separate reachability check.

Worked examples

Example 1 (same base). Simplify 25×4382\frac{2^5 \times 4^3}{8^2}.

  • 43=(22)3=264^3 = (2^2)^3 = 2^6 and 82=(23)2=268^2 = (2^3)^2 = 2^6.
  • 25×2626=25=32\frac{2^5 \times 2^6}{2^6} = 2^5 = 32.

Example 2 (unknown in the exponent). Solve 32x−1=2433^{2x - 1} = 243.

  • 243=35243 = 3^5, so 2x−1=52x - 1 = 5 and x=3x = 3.

Example 3 (fractional and negative exponents). 272/3=(273)2=32=927^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9, and 16−3/4=1/(164)3=1/23=1/816^{-3/4} = 1 / (\sqrt[4]{16})^3 = 1 / 2^3 = 1/8.

Example 4 (comparison). Which is larger, 23002^{300} or 32003^{200}?

  • 2300=(23)100=81002^{300} = (2^3)^{100} = 8^{100} and 3200=(32)100=91003^{200} = (3^2)^{100} = 9^{100}.
  • Equal exponents, larger base wins: 32003^{200} is larger.

Example 5 (units digit). Find the units digit of 71237^{123}.

  • The cycle for 7 is 7, 9, 3, 1 (period 4).
  • 123=4×30+3123 = 4 \times 30 + 3, so take the 3rd term of the cycle: the units digit is 3.

Example 6 (number of digits). How many digits does 21002^{100} have? Use log⁡102≈0.30103\log_{10} 2 \approx 0.30103.

  • log⁡102100=100×0.30103=30.103\log_{10} 2^{100} = 100 \times 0.30103 = 30.103.
  • Digits: ⌊30.103⌋+1=31\lfloor 30.103 \rfloor + 1 = 31.

Example 7 (log arithmetic). Evaluate log⁡28+log⁡381−log⁡5125\log_2 8 + \log_3 81 - \log_5 125.

  • log⁡28=3\log_2 8 = 3, log⁡381=4\log_3 81 = 4, log⁡5125=3\log_5 125 = 3, so the value is 3+4−3=43 + 4 - 3 = 4.

Example 8 (building a log from given values). Given log⁡102=0.3010\log_{10} 2 = 0.3010 and log⁡103=0.4771\log_{10} 3 = 0.4771, find log⁡1072\log_{10} 72.

  • 72=23×3272 = 2^3 \times 3^2, so log⁡1072=3(0.3010)+2(0.4771)=0.9030+0.9542=1.8572\log_{10} 72 = 3(0.3010) + 2(0.4771) = 0.9030 + 0.9542 = 1.8572.
  • With more decimal places the true value is 1.857331.85733 to five places; the four-place inputs give the answer to within 0.0002.

Example 9 (change of base). log⁡432=log⁡232log⁡24=52\log_4 32 = \frac{\log_2 32}{\log_2 4} = \frac{5}{2}.

Example 10 (logarithmic equation with a rejected root). Solve log⁡10x+log⁡10(x−3)=1\log_{10} x + \log_{10}(x - 3) = 1.

  • Combine: log⁡10(x(x−3))=1\log_{10}(x(x - 3)) = 1, so x2−3x=10x^2 - 3x = 10, that is x2−3x−10=0x^2 - 3x - 10 = 0, giving x=5x = 5 or x=−2x = -2.
  • x=−2x = -2 makes log⁡10x\log_{10} x undefined, so the only solution is x=5x = 5.

Example 11 (growth to a target). Money grows at 8% per year, compounded annually. After how many whole years has it first at least doubled?

  • n=ln⁡2/ln⁡1.08≈9.006n = \ln 2 / \ln 1.08 \approx 9.006, so the ceiling is 10.
  • Check by powers: 1.089≈1.9991.08^9 \approx 1.999 (just short of 2) and 1.0810≈2.1591.08^{10} \approx 2.159. The answer is 10 years. Rounding the logarithm to 9 gives the wrong option.

Example 12 (surds). 50+18=52+32=82\sqrt{50} + \sqrt{18} = 5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}, and 13−2=3+23−2=3+2\frac{1}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} + \sqrt{2}}{3 - 2} = \sqrt{3} + \sqrt{2} (multiply by the conjugate).

Important properties and observations

  • Common misconceptions. ln⁡(a+b)≠ln⁡a+ln⁡b\ln(a + b) \ne \ln a + \ln b; (a+b)2≠a2+b2(a + b)^2 \ne a^2 + b^2; aman≠amna^{m} a^{n} \ne a^{mn}; log⁡ab⋅log⁡ba=1\log_a b \cdot \log_b a = 1 (not log⁡a(ab)\log_a (ab)).
  • Tower convention. abca^{b^c} means a(bc)a^{(b^c)}, which differs from (ab)c=abc(a^b)^c = a^{bc}: 232=29=5122^{3^2} = 2^9 = 512 but (23)2=64(2^3)^2 = 64.
  • Monotonicity. For base a>1a > 1, log⁡a\log_a is increasing, so inequalities survive taking logs; for 0<a<10 < a < 1 they reverse.
  • Logs of numbers near powers of 10. log⁡10\log_{10} of a number between 1 and 10 lies between 0 and 1; multiplying the number by 10k10^k adds kk to its base-10 logarithm (a rightward decimal shift by kk places for positive kk); division by 10k10^k subtracts kk.
  • Useful values. log⁡102≈0.3010\log_{10} 2 \approx 0.3010, log⁡103≈0.4771\log_{10} 3 \approx 0.4771, log⁡105=1−log⁡102≈0.6990\log_{10} 5 = 1 - \log_{10} 2 \approx 0.6990, log⁡107≈0.8451\log_{10} 7 \approx 0.8451, 210=1024≈1032^{10} = 1024 \approx 10^3.
  • EDGE CASE. The base of a logarithm can never be 1, and the argument must be positive; candidates that break this are rejected (Example 10).
  • Related material. The "rule of 70" estimates the doubling time at r%r\% growth as about 70/r70 / r periods (from ln⁡2≈0.693\ln 2 \approx 0.693); at 8% it suggests about 8.75 years, which shows why an estimate must be confirmed by powers when the question asks for whole periods.

GATE examination perspective

  • Base-rewriting questions. Expressions with 4, 8, 16, 32 or 9, 27, 81 collapse to one base when rewritten (Example 1).
  • Exponent-equation traps. After equating exponents, solve the linear equation carefully; a sign slip in 2x−12x - 1 is a typical distractor (Example 2).
  • Comparison traps. The larger exponent does not decide by itself; equalise exponents or compare nln⁡an \ln a (Example 4).
  • Units-digit questions. Handle exponent 0 separately; for positive integer exponents, reduce the exponent modulo 4, and use the 4th term when the remainder is 0.
  • Digit-count questions. Use ⌊log⁡10N⌋+1\lfloor \log_{10} N \rfloor + 1, and remember that 10k10^k itself has k+1k + 1 digits.
  • Off-by-one in growth. Ceil the exact quotient and verify with the neighbouring powers (Example 11).
  • Domain traps. Logarithmic equations often produce an extraneous root (Example 10).

Summary

Sources

  1. J. Abramson et al., College Algebra 2e, OpenStax, 2021: Section 1.2 Exponents and Scientific Notation; Section 1.3 Radicals and Rational Exponents; Sections 6.3 Logarithmic Functions, 6.5 Logarithmic Properties, 6.6 Exponential and Logarithmic Equations. (consulted 2026-11-11).
  2. D. Kirk et al., Contemporary Mathematics, OpenStax, 2023: Section 3.8 Exponents. (consulted 2026-11-11).

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