NDA Maths · Logarithms

Logarithm Identities, Change of Base & Sums

A logarithm answers “what power do I raise the base to?” — and almost every NDA log question collapses once you apply the three laws, the change-of-base rule, or its reciprocal twin.

Why this matters

This is the larger of the chapter's two subtopics — 16 PYQs, mostly EASY/MODERATE with 2 HARD. The patterns repeat: split or combine logs with the laws, switch base to collapse a product or a telescoping sum (the 1/log_k N family appears almost every other year), or read off the sign of a log. Master change of base and these become one-liners.

Concept 1 of 5

What a Logarithm Is — Laws, Special Values, Domain

Intuition

A logarithm is just an exponent turned inside out. The question “logaN=?\log_a N = ?” asks “to what power must I raise the base aa to get NN?” Once you read every log as “the exponent that produces NN,” the three laws are simply the exponent rules in disguise.

Definition

Definition. logaN=x    ax=N\log_a N = x \iff a^x = N, valid for base a>0, a1a>0,\ a\neq 1 and N>0N>0. The three laws (same base throughout):

  • Product: loga(MN)=logaM+logaN\log_a(MN) = \log_a M + \log_a N
  • Quotient: loga ⁣MN=logaMlogaN\log_a\!\dfrac{M}{N} = \log_a M - \log_a N
  • Power: loga(Mk)=klogaM\log_a(M^k) = k\,\log_a M

Special values: logaa=1\log_a a = 1, loga1=0\log_a 1 = 0, loga(ak)=k\log_a(a^k) = k, and alogaN=Na^{\log_a N} = N. Domain: you may only take the log of a positive number — the argument of every log in a problem must stay >0>0. This is what later forces solution-rejection.

The defining equivalence and the three laws

logaN=x    ax=N;loga(MN)=logaM+logaN,  logaMk=klogaM\log_a N = x \iff a^x = N;\quad \log_a(MN)=\log_a M+\log_a N,\ \ \log_a M^k = k\log_a M

Worked example

Evaluate log280log25\log_2 80 - \log_2 5.
Practice this concept4 quick reps

log(M+N)\log(M+N) is NOT logM+logN\log M + \log N

The product law splits a log of a product, not a log of a sum. loga(M+N)\log_a(M+N) has no simplification — only loga(MN)\log_a(MN) splits into logaM+logaN\log_a M+\log_a N.

The base must be >0>0 and 1\neq 1

logaN\log_a N is only defined for a base a>0, a1a > 0,\ a \neq 1. A base of 11 is forbidden because 1x=11^x = 1 for every xx, so log1N\log_1 N has no unique value; a base 0\le 0 breaks the exponential entirely.

Concept 2 of 5

Applying the Laws — Evaluate and Combine

Intuition

Most “find the value” log questions are solved by rewriting every number as a power of a small base, then letting the power and product/quotient laws do the bookkeeping. The trick is to spot the common base hiding inside 27,32,102427, 32, 1024, nested radicals, and so on.

Definition

Strategy for an evaluate-and-combine problem:

  • Rewrite arguments as powers of the smallest convenient base: 27=33, 1024=210, 3125=5527=3^3,\ 1024=2^{10},\ 3125=5^5, and a nested radical like 777\sqrt{7\sqrt{7\sqrt{7}}} as 71/2+1/4+1/8=77/87^{1/2+1/4+1/8}=7^{7/8}.
  • Pull exponents out front with the power law, then combine with product/quotient.
  • Sum of logs across a list — e.g. jlog10(2j5j)=jjlog1010=jj\sum_j \log_{10}(2^j5^j)=\sum_j j\,\log_{10}10=\sum_j j — telescopes once you notice 25=102\cdot 5 = 10.

Rewrite as powers, then pull the exponent out

loga(bmcn)=mlogab+nlogac\log_a(b^m \cdot c^n) = m\log_a b + n\log_a c

Worked example

Find the value of log327+log5125\log_3 27 + \log_5 125.
Practice this conceptself-check · 1 quick reps

From the bank · past-year question

Example 2LogarithmsMODERATE
What is the value of log7log7777\log_7 \log_7 \sqrt{7\sqrt{7\sqrt{7}}} equal to?

[Q1 · Sep · 2018]

Keep the base when you pull out a power

log7777\log_7\sqrt{7\sqrt{7\sqrt{7}}} first becomes 78\tfrac{7}{8}, but a second outer log7\log_7 of that gives log778=13log72\log_7\tfrac{7}{8} = 1 - 3\log_7 2 — don't lose the log78=3log72-\log_7 8 = -3\log_7 2 term.

Concept 3 of 5

Change of Base & the Reciprocal Identity

Intuition

Any logarithm can be re-expressed in a base of your choosing: logba=logalogb\log_b a = \dfrac{\log a}{\log b} in any common base. The single most useful consequence is that flipping a log inverts it: 1logab=logba\dfrac{1}{\log_a b} = \log_b a. That reciprocal turns the recurring 1logkN\sum \dfrac{1}{\log_k N} sums into a one-line collapse.

Definition

Change of base: logba=logcalogcb\log_b a = \dfrac{\log_c a}{\log_c b} for any valid base cc. Two consequences carry most of the marks:

  • Reciprocal identity: 1logab=logba\dfrac{1}{\log_a b} = \log_b a (set c=ac=a). So 1logkN=logNk\dfrac{1}{\log_k N} = \log_N k.
  • Telescoping sum: k=2m1logkN=k=2mlogNk=logN(23m)=logN(m!)\displaystyle\sum_{k=2}^{m}\dfrac{1}{\log_k N} = \sum_{k=2}^{m}\log_N k = \log_N(2\cdot 3\cdots m) = \log_N(m!). When N=m!N=m! this equals logNN=1\log_N N = 1.
  • Product collapse: logablogba=1\log_a b\cdot\log_b a = 1, and logablogbc=logac\log_a b\cdot\log_b c = \log_a c (chain).

Change of base and its reciprocal twin

logba=logalogb,1logab=logba\log_b a = \dfrac{\log a}{\log b}, \qquad \dfrac{1}{\log_a b} = \log_b a

Worked example

If n=50!n = 50!, find 1log2n+1log3n++1log50n\dfrac{1}{\log_2 n} + \dfrac{1}{\log_3 n} + \cdots + \dfrac{1}{\log_{50} n}.
Practice this conceptself-check · 1 quick reps

From the bank · past-year question

Example 3LogarithmsMODERATE
If n=(2017)!n = (2017)!, then what is 1log2n+1log3n+1log4n++1log2017n\frac{1}{\log_{2}n} + \frac{1}{\log_{3}n} + \frac{1}{\log_{4}n} + \cdots + \frac{1}{\log_{2017}n} equal to?

[Q2 · Apr · 2018]

Reciprocal flips the base and the argument together

1logab=logba\dfrac{1}{\log_a b} = \log_b a — the base and argument swap. It does NOT equal loga(1/b)\log_a(1/b) (which would be logab-\log_a b). Keep the two operations separate.

Change of base does NOT invert the fraction

logba=logalogb\log_b a = \dfrac{\log a}{\log b} — the NEW argument aa goes on top, the NEW base bb on the bottom. Writing it upside-down as logbloga\dfrac{\log b}{\log a} gives logab\log_a b, the reciprocal, and flips the whole answer.

Concept 4 of 5

Sign of a Logarithm & Bounds of a Log Function

Intuition

Because the log of 11 is 00, a logarithm is positive when its argument exceeds 11 and negative when the argument lies strictly between 00 and 11 (for a base >1>1). And since log\log is increasing, the smallest value of log10(quadratic)\log_{10}(\text{quadratic}) happens exactly where the quadratic is smallest.

Definition

For base a>1a>1:

  • logaN>0    N>1\log_a N > 0 \iff N > 1; logaN=0    N=1\quad\log_a N = 0 \iff N = 1; logaN<0    0<N<1\quad\log_a N < 0 \iff 0 < N < 1.
  • The function loga\log_a is strictly increasing, so logaf(x)\log_a f(x) attains its minimum exactly where f(x)f(x) is minimised (provided f>0f>0 there).

**Minimising log10(quadratic):\log_{10}(\text{quadratic}):** complete the square — x2+bx+c=(x+b2)2+(cb24)x^2+bx+c = (x+\tfrac{b}{2})^2 + (c-\tfrac{b^2}{4}) — the minimum argument is cb24c-\tfrac{b^2}{4}, and the minimum of the log is log10\log_{10} of that.

Sign of a log (base > 1)

logaN  {>0N>1=0N=1<00<N<1\log_a N \;\begin{cases}>0 & N>1\\ =0 & N=1\\ <0 & 0<N<1\end{cases}

Worked example

Find the minimum value of f(x)=log10(x24x+104)f(x) = \log_{10}(x^2 - 4x + 104).
Practice this concept2 quick reps

From the bank · past-year question

Example 4LogarithmsEASY
What is the minimum value of the function f(x)=log10(x2+2x+11)f(x) = \log_{10}(x^2+2x+11)?

[Q73 · Sep · 2022]

The minimum is of the LOG, not the quadratic

After finding the quadratic's minimum (say 100100), you must still take log10100=2\log_{10} 100 = 2. The smallest argument and the smallest log value are different numbers — the question asks for the log.

Concept 5 of 5

Logarithms in AP/GP and the Geometric Mean

Intuition

Logs convert multiplication into addition, so a geometric progression of powers becomes an arithmetic progression of logs. That's why questions phrased as “lnx,lnx3,lnx5\ln x, \ln x^3, \ln x^5 are in AP/GP?” or “geometric mean of 1,2,22,1,2,2^2,\dots” are pure log-law exercises in disguise.

Definition

Two recurring shapes:

  • AP / GP test on logs: lnx,lnx3,lnx5=lnx,3lnx,5lnx\ln x, \ln x^3, \ln x^5 = \ln x,\,3\ln x,\,5\ln x. They are in AP (common difference 2lnx2\ln x); for GP you must separately check q2=prq^2 = pr, i.e. (3lnx)2=(lnx)(5lnx)(3\ln x)^2 = (\ln x)(5\ln x) — here 959 \neq 5, so never GP.
  • Geometric mean of a list of powers: the GM of 1,2,22,,2n11,2,2^2,\dots,2^{n-1} is (20+1++(n1))1/n=2(n1)/2\big(2^{0+1+\cdots+(n-1)}\big)^{1/n} = 2^{(n-1)/2}; taking log2\log_2 gives n12\dfrac{n-1}{2}, so expressions like 1+2log2G1+2\log_2 G simplify to nn.

AP and GP conditions for three terms

AP:2q=p+r,GP:q2=pr\text{AP}: 2q = p+r, \qquad \text{GP}: q^2 = pr

Worked example

Are log24, log216, log264\log_2 4,\ \log_2 16,\ \log_2 64 in AP?
Practice this conceptself-check

From the bank · past-year question

Example 5LogarithmsMODERATE
Let p=lnxp=\ln x, q=lnx3q=\ln x^3 and r=lnx5r=\ln x^5, where x>1x>1. Which of the following statements is/are correct? A. p,qp,q and rr are in AP. B. p,qp,q and rr can never be in GP. Select the answer using the code given below.

[Q8 · Sep · 2024]

AP holding does not make it GP — test GP separately

p,q,rp, q, r being in AP says nothing about GP. The GP condition q2=prq^2 = pr must be checked on its own; for lnx,3lnx,5lnx\ln x, 3\ln x, 5\ln x it fails because 959 \neq 5, so the terms are AP-but-never-GP.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

  • What a Logarithm Is — Laws, Special Values, Domain

    The defining equivalence and the three laws

    logaN=x    ax=N;loga(MN)=logaM+logaN,  logaMk=klogaM\log_a N = x \iff a^x = N;\quad \log_a(MN)=\log_a M+\log_a N,\ \ \log_a M^k = k\log_a M
  • Applying the Laws — Evaluate and Combine

    Rewrite as powers, then pull the exponent out

    loga(bmcn)=mlogab+nlogac\log_a(b^m \cdot c^n) = m\log_a b + n\log_a c
  • Change of Base & the Reciprocal Identity

    Change of base and its reciprocal twin

    logba=logalogb,1logab=logba\log_b a = \dfrac{\log a}{\log b}, \qquad \dfrac{1}{\log_a b} = \log_b a
  • Sign of a Logarithm & Bounds of a Log Function

    Sign of a log (base > 1)

    logaN  {>0N>1=0N=1<00<N<1\log_a N \;\begin{cases}>0 & N>1\\ =0 & N=1\\ <0 & 0<N<1\end{cases}
  • Logarithms in AP/GP and the Geometric Mean

    AP and GP conditions for three terms

    AP:2q=p+r,GP:q2=pr\text{AP}: 2q = p+r, \qquad \text{GP}: q^2 = pr

Watch out for (7)

Drill every past-year question on this subtopic

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