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GMAT Focus Edition · Advanced Quantitative Aptitude

Logarithms, Surds and Indices

Laws of exponents, radicals, and logarithmic identities and manipulation.

Six concepts — definition and domain first, then the log laws and change of base that turn every equation into index arithmetic, the index laws those logs mirror, digit counting, and the two surd moves CAT actually asks for. Nothing here needs cleverness; it needs the rules applied in the right order.

  • GMAT Focus Edition
  • Medium level
  • 6 concepts
  • 21 practice questions

1What a logarithm means — and where it stops

\log_b a = x means b^x = a: the logarithm is the exponent that raises the base to the argument. In CAT you almost never compute that exponent by hand; you rewrite the statement, combine logs, or change base. Before any of that, lock the domain: the base satisfies b > 0 with b \ne 1, and the argument satisfies a > 0. A quadratic found after combining logs can still produce roots that violate a > 0 — those roots look like algebra but are not solutions.

Figure. \log_2 x = y means 2^y = x. The curves are mirrors across y=x. Domain of \log_b needs x>0 and base b>0, b\neq 1.

How it works

  1. Read the definition\log_b a = x and b^x = a say the same thing; pick whichever form exposes the exponent.
  2. State the domain firstRequire b > 0, b \ne 1, and a > 0 before manipulating — each log you combine inherits its argument's restriction.
  3. Reject extraneous rootsAfter solving, every candidate must satisfy every log's argument inequality; discard anything that does not.
Domain conditions (real logarithms)
SymbolMust satisfyIf violated
Base bb > 0b^x is not defined as a real power for non-positive b
Base bb \ne 1\log_1 a is undefined for every a \ne 1
Argument aa > 0No real \log of zero or a negative number

Solving a logarithmic equation

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

  • Combine logs: \log_2\big(x(x-2)\big) = 3x(x-2) = 2^3 = 8
  • Quadratic: x^2 - 2x - 8 = 0(x-4)(x+2) = 0
  • Domain needs x > 2; reject x = -2x = 4

Pro tip. Always check candidate solutions against every log's domain; a negative or too-small root is silently invalid and CAT options often include it.

Solve \log_3(x-1) = 2.
  1. x = 10
  2. x = 8
  3. x = -8

3^2 = 9 so x - 1 = 9 and x = 10. The argument must stay positive, so x > 1 — 10 passes. 8 comes from forgetting to add 1 after reading x - 1 = 9; -8 is the sign slip that also fails the domain.

2The three log laws — product, quotient, power

Every log manipulation in CAT is one of three index laws read backwards. A product inside the log becomes a sum: \log(mn) = \log m + \log n. A quotient becomes a difference: \log\frac{m}{n} = \log m - \log n. A power brings the exponent down as a multiplier: \log m^k = k\log m. Mismatched bases are not fixed by these laws — you must change base first. The trap is applying the product law to a sum — log of a sum is never the sum of logs.

Figure. Three laws, one idea: logs turn product into sum, quotient into difference, and power into a multiplier in front. Same base on every term.

How it works

  1. Spot the structureMultiplication or division inside one log → combine to a single log. A power on the argument → pull the exponent out front.
  2. Same base throughoutEach term must already share the same base before you add or subtract logs; otherwise change base first.
  3. Reverse when solvingAn equation like \log x + \log(x-3) = 2 becomes \log\big(x(x-3)\big) = 2, then x(x-3) = b^2 — the laws run in both directions.

Combining two logs

Evaluate \log_2 32 + \log_2 2.

  • Product law: \log_2 32 + \log_2 2 = \log_2(32 \times 2)\log_2 64
  • Read the exponent: 2^x = 64x = 6
  • Check directly: \log_2 32 = 5, \log_2 2 = 15 + 1 = 6

Pro tip. When several logs share a base, combine them into one log before evaluating — it is faster than evaluating each term separately and it is the move every log equation expects.

Which expression equals \log_3 81?
  1. 4
  2. 3
  3. 27

3^4 = 81, so \log_3 81 = 4. The value 3 is the base, not the log; 27 is 3^3, which would make \log_3 27 = 3.

3Change of base — and the reciprocal shortcut

Different bases in one equation cannot be combined until they match. Change of base rewrites any log on a convenient common base: \log_b a = \frac{\log_c a}{\log_c b} for any valid c. On CAT, c = 10 is the default because \log_{10} 2 \approx 0.3010 and \log_{10} 3 \approx 0.4771 are worth memorising — from them you build \log_{10} 4, \log_{10} 5, \log_{10} 6, \log_{10} 8 and \log_{10} 9 using the product and quotient laws. The reciprocal identity \log_a b \cdot \log_b a = 1 collapses chained logs instantly.

Figure. Change of base: log_b a = log_c a / log_c b for any valid c. The reciprocal shortcut is the same identity with c = a, so log_a b · log_b a = 1. Given log10 2 = 0.3010, log10 5 = log10(10/2) = 1 − 0.3010 = 0.6990.

How it works

  1. Pick a common baseRewrite every log as a ratio of base-10 logs (or natural logs if the problem supplies \ln).
  2. Build unknown logs from known ones\log_{10} 5 = \log_{10} 10 - \log_{10} 2 = 1 - 0.3010; \log_{10} 6 = \log_{10} 2 + \log_{10} 3.
  3. Use reciprocals on chainsWhen \log_a b and \log_b a appear together, their product is 1 — no conversion needed.

Finding \log_{10} 5 from \log_{10} 2

Given \log_{10} 2 = 0.3010, find \log_{10} 5.

  • Write 5 = \frac{10}{2}, so \log_{10} 5 = \log_{10} 10 - \log_{10} 21 - 0.3010
  • Subtract0.6990
  • Check via change of base: \log_2 5 = \frac{0.6990}{0.3010}\approx 2.32, and 2^{2.32} \approx 5

Pro tip. Convert every logarithm to a single base before solving an equation — mismatched bases cause most mistakes, and the reciprocal rule saves a conversion when the same two numbers appear as argument and base in swapped roles.

If \log_a b = 3, then \log_b a equals
  1. \frac{1}{3}
  2. 3
  3. -3

\log_a b \cdot \log_b a = 1, so \log_b a = 1/3. The value 3 repeats the given log instead of inverting it; -3 confuses reciprocal with negation.

4Index laws — the engine behind every log step

Logarithms are exponents read backwards, so every log law has an index counterpart. Multiply powers of the same base and add exponents: a^m \cdot a^n = a^{m+n}. Divide and subtract: \frac{a^m}{a^n} = a^{m-n}. Raise a power to a power and multiply exponents: (a^m)^n = a^{mn}. Any non-zero base to the power zero is one. A negative exponent means reciprocal: a^{-n} = 1/a^n. Fractional exponents are roots: a^{m/n} = \sqrt[n]{a^m}. The inverse identity a^{\log_a x} = x is the bridge back from logs to ordinary algebra.

Keep the mirror in view: b^m\cdot b^n=b^{m+n} pairs with \log_b(uv)=\log_b u+\log_b v, and (b^m)^n=b^{mn} pairs with \log_b(u^n)=n\log_b u. Negative and fractional indices are still these laws.

How it works

  1. Same base before combiningIndex laws apply only when the bases match; rewrite 4^x as (2^2)^x = 2^{2x} when pairing it with a power of 2.
  2. Add exponents on multiplya^m \cdot a^n = a^{m+n} — the move that becomes the product law for logs.
  3. Fractional powers are rootsa^{1/2} = \sqrt{a} and a^{2/3} = \sqrt[3]{a^2}; convert surds to fractional exponents when simplifying products.
Index laws at a glance
OperationRuleLog mirror
Multiplya^m \cdot a^n = a^{m+n}\log(mn) = \log m + \log n
Divide\frac{a^m}{a^n} = a^{m-n}\log\frac{m}{n} = \log m - \log n
Power of a power(a^m)^n = a^{mn}\log m^k = k\log m
Zero exponenta^0 = 1 (a \ne 0)\log_a 1 = 0

Multiplying fractional powers

Simplify 2^{1/2} \times 2^{3/2}.

  • Same base — add exponents: 2^{1/2 + 3/2}2^{4/2} = 2^2
  • Evaluate4
  • Check: 2^{1/2} = \sqrt{2}, 2^{3/2} = 2\sqrt{2}, product = 2 \times 2 = 44

Pro tip. When a problem mixes surds and powers, rewrite every root as a fractional exponent first — addition of exponents beats rationalising mid-stream.

Which equals (3^2)^3?
  1. 3^6
  2. 3^5
  3. 3^8

(a^m)^n = a^{mn}, so (3^2)^3 = 3^{2 \times 3} = 3^6. The value 3^5 adds exponents instead of multiplying them; 3^8 confuses a power tower 3^{(2^3)} with the power-of-a-power rule.

5Counting digits with base-10 logs

A positive integer N has \lfloor \log_{10} N \rfloor + 1 digits. The log tells you how far N sits above the nearest power of ten: if 10^k \le N < 10^{k+1}, then k \le \log_{10} N < k+1, so the floor is k and adding one counts every digit place from 10^0 through 10^k. For a power like 2^{50}, compute \log_{10}(2^{50}) = 50 \log_{10} 2 — you never need the full number. CAT supplies \log_{10} 2 \approx 0.3010 when it wants this route.

Figure. The characteristic of \log_{10} n is \lfloor\log_{10} n\rfloor; add one for the number of decimal digits. Here 10^3 \le 1234 < 10^4 so four digits.

How it works

  1. Take the base-10 logFor N = a^b, use \log_{10}(a^b) = b \log_{10} a.
  2. Take the floor\lfloor \log_{10} N \rfloor is the exponent of the highest power of ten not exceeding N.
  3. Add oneThe digit count is that floor plus one — do not skip the +1; \lfloor \log_{10} 999 \rfloor = 2 but 999 has three digits.

How many digits does 2^{50} have?

How many digits does 2^{50} have? (Use \log_{10} 2 = 0.3010.)

  • \log_{10}(2^{50}) = 50 \times 0.301015.05
  • \lfloor 15.05 \rfloor + 115 + 1 = 16
  • Sanity: 2^{10} has 4 digits; each further +10 in the exponent adds about 3 digit places, so 2^{50} is +40 in the exponent \approx +12 digits4 + 12 = 16 digits

Pro tip. Take the floor of the base-10 log and add one — this instantly gives the digit count of any large power without evaluating the power itself.

If \log_{10} 2 = 0.3, how many digits does 2^{10} have?
  1. 4
  2. 3
  3. 5

\log_{10}(2^{10}) = 10 \times 0.3 = 3, so \lfloor 3 \rfloor + 1 = 4 digits. Indeed 2^{10} = 1024. The answer 3 forgets the +1; 5 treats the log itself as the digit count.

6Nested surds and rationalising denominators

A surd is an irrational root left in exact form — \sqrt{2}, \sqrt{3}, and nested forms like \sqrt{7 + 2\sqrt{10}}. To denest, assume \sqrt{a + 2\sqrt{b}} = \sqrt{x} + \sqrt{y} and match x + y = a with xy = b — the pattern comes from (\sqrt{x} + \sqrt{y})^2 = x + y + 2\sqrt{xy}. To rationalise a denominator, multiply numerator and denominator by the conjugate: (\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b clears the radical from the bottom. Both moves keep the value exact while making the expression easier to compare or add.

Figure. Denest √(7+2√10) by matching x+y=7 and xy=10, so the surd is √5+√2. Rationalise 1/(√3−1) by multiplying top and bottom by the conjugate √3+1; the denominator becomes 3−1=2, and the value is (√3+1)/2.

How it works

  1. Denest by matching sum and productFor \sqrt{a + 2\sqrt{b}}, find two positive numbers with sum a and product b; then the surd equals their root sum.
  2. Rationalise with the conjugateFor \frac{1}{\sqrt{p} - \sqrt{q}}, multiply top and bottom by \sqrt{p} + \sqrt{q} so the denominator becomes p - q.
  3. Simplify the integer denominatorAfter rationalising, cancel any common factor between the surd numerator and the integer denominator.

Denesting and rationalising

Simplify \sqrt{7 + 2\sqrt{10}} and \dfrac{1}{\sqrt{3} - 1}.

  • Denest: try \sqrt{x} + \sqrt{y} with x + y = 7, xy = 10x = 5, y = 2
  • So \sqrt{7 + 2\sqrt{10}} = \sqrt{5} + \sqrt{2}\sqrt{5} + \sqrt{2}
  • Rationalise \frac{1}{\sqrt{3}-1}: multiply by \frac{\sqrt{3}+1}{\sqrt{3}+1}\dfrac{\sqrt{3} + 1}{2}

Pro tip. Match a nested surd to (\sqrt{x} + \sqrt{y})^2 by finding two numbers with the right sum and product; for a single-surd denominator, the conjugate always produces a clean difference of squares.

Rationalising \dfrac{1}{\sqrt{5} - 2} gives
  1. \sqrt{5} + 2
  2. \dfrac{\sqrt{5} + 2}{3}
  3. \dfrac{\sqrt{5} - 2}{3}

Multiply by \frac{\sqrt{5}+2}{\sqrt{5}+2}: the denominator is 5 - 4 = 1, so the value is just \sqrt{5} + 2. The fraction with denominator 3 would come from \frac{1}{\sqrt{5}-2} only if the denominator were \sqrt{5}-2 paired incorrectly; here 5 - 4 = 1, not 3.

Notes

  • Log definition: \log_b a=x means b^x=a, which requires a>0, b>0, and b\ne1.
  • Laws of logarithms: \log(mn)=\log m+\log n, \log\frac{m}{n}=\log m-\log n, and \log m^k=k\log m.
  • Change of base: \log_b a=\frac{\log_c a}{\log_c b}, which lets any logarithm be re-expressed on a convenient common base.
  • Laws of indices: a^m\cdot a^n=a^{m+n}, (a^m)^n=a^{mn}, and a^0=1 for any a\ne0.
  • Rationalizing surds: multiplying by the conjugate works because (\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=a-b clears the radical from the denominator.

Formulas

  • Change of base: \log_b a=\frac{\ln a}{\ln b}
  • Power rule: \log_b(a^k)=k\log_b a
  • Reciprocal rule: \log_b a=\frac{1}{\log_a b}
  • Product of indices: (a^m)^n=a^{mn}
  • Fractional exponent: a^{m/n}=\sqrt[n]{a^m}
  • Inverse identity: a^{\log_a x}=x

Exam traps & shortcuts

  • The number of digits in a positive integer N is \lfloor\log_{10}N\rfloor+1.
  • Use \log_a b\cdot\log_b a=1 to cancel chained logarithms quickly.
  • Convert every logarithm to a single base before solving an equation; mismatched bases cause most mistakes.
  • For a nested surd \sqrt{a+2\sqrt{b}}, try \sqrt{x}+\sqrt{y} and match x+y=a with xy=b.

Reference tables

These appear across the chapter — change of base, the reciprocal rule, and the bridge between logs and indices.

Formulas worth memorising
NameFormula
Change of base\log_b a = \dfrac{\ln a}{\ln b}
Power rule\log_b(a^k) = k\log_b a
Reciprocal rule\log_b a = \dfrac{1}{\log_a b}
Product of indices(a^m)^n = a^{mn}
Fractional exponenta^{m/n} = \sqrt[n]{a^m}
Inverse identitya^{\log_a x} = x

Recap

Read only this the night before.

Domain
a > 0, b > 0, b \ne 1. After solving, reject any root that violates an argument inequality — CAT options often include extraneous values.
Log laws
Product → sum, quotient → difference, power → multiplier. Never split \log(m+n) into a sum of logs.
Change of base
Convert every log to one base before combining. \log_{10} 2 \approx 0.3010 and \log_{10} 3 \approx 0.4771 build the rest; \log_a b \cdot \log_b a = 1.
Indices
Same-base powers add exponents on multiply, subtract on divide, multiply on power-of-a-power. Fractional exponents are roots.
Digits
Digit count = \lfloor \log_{10} N \rfloor + 1. For a^b, compute b \log_{10} a — do not evaluate a^b.
Surds
Denest via x+y and xy; rationalise with the conjugate so the denominator becomes a difference of squares.

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