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Indices, logarithms & surds · Lesson 2 of 3

Logarithms at Further Maths depth

Simplifying awkward logs without tables, changing the base (and why logₓ a = 1 ÷ logₐ x), and solving log equations, including pairs of them.

18 minYou should already know: Indices & standard form Logarithms Surds
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A logarithm is an index. log⁡ax=n\log_a x = n says the same as an=xa^n = x, and the three laws come straight from the laws of indices (see logarithms):

log₂ 32 = 5means2⁵ = 32basepower
The log is the powerlogₐ x = n means aⁿ = x
log⁡a(xy)=log⁡ax+log⁡aylog⁡axy=log⁡ax−log⁡aylog⁡axn=nlog⁡ax\begin{aligned} \log_a(xy) &= \log_a x + \log_a y \\ \log_a\frac{x}{y} &= \log_a x - \log_a y \\ \log_a x^n &= n\log_a x \end{aligned}

Further Maths uses these on harder numbers, adds one new rule (changing the base), and asks you to solve equations with them. The laws are only for the log of a product, a quotient or a power: log⁡(x+y)\log(x + y) can’t be split.

Simplifying without tables

Break every number into prime powers, including surds, since 7=712\sqrt7 = 7^{\frac12}. Then each log becomes a multiple of log⁡\log of a prime, and the answer usually cancels to a plain number.

Worked example · WAEC 2012

WAEC 2012 · Paper 2 · Q1

Without using tables or calculators, simplify log⁡77−log⁡55log⁡1.4\dfrac{\log7\sqrt7 - \log5\sqrt5}{\log1.4}.

  1. Write the surds as powers

    • 77=71×712=732{7\sqrt7 = 7^1 \times 7^{\frac12} = 7^{\frac32}} (add the indices).
    • In the same way, 55=532{5\sqrt5 = 5^{\frac32}}.

    Think first. 7√7 is 7 times 7 to the half. What single power of 7 is that?

  2. Use the power law on top

    • Bring each power down in front: log⁡732−log⁡532=32log⁡7−32log⁡5{\log 7^{\frac32} - \log 5^{\frac32} = \frac32\log 7 - \frac32\log 5}.
    • Take out the common 32\frac32: the top is 32(log⁡7−log⁡5)\frac32(\log 7 - \log 5).

  3. Write the bottom the same way

    • 1.4=1410=75{1.4 = \frac{14}{10} = \frac75}.
    • So log⁡1.4=log⁡7−log⁡5{\log 1.4 = \log 7 - \log 5}.

    Think first. 1.4 is a fraction. Which one?

  4. Divide

    • The bracket (log⁡7−log⁡5)(\log 7 - \log 5) is on the top and the bottom, so it cancels.
    • The answer is 32\dfrac32.

More: simplifying without tables

Changing the base

A calculator only has log⁡\log (base 10). To find a log in any other base, divide:

log⁡ax=log⁡xlog⁡a\log_a x = \frac{\log x}{\log a}

The same rule with the base and the number swapped gives a second fact, which turns up in equations: log⁡xa=1log⁡ax\log_x a = \dfrac{1}{\log_a x}.

logₐ x = log x ÷ log alog₉ 27 = log 27 ÷ log 9= 3 log 3 ÷ 2 log 3 = 3⁄2logₓ a = 1 ÷ logₐ xswap base and number: the log turns over
Change of baseDivide by the log of the base; swap them and the log turns upside down
Change of basePick a base, slide x
48121620242832−2−1123456xy(8, 1.5)y = log₄ x
0.9031 ÷ 0.6021log 8 ÷ log 41.5log₄ 88check: 4 to this power
Any base can be changed to base 10: log₄ 8 = log 8 ÷ log 4 = 1.5. The check raises 4 to that power and gets 8 back (up to rounding). The answer does not have to be a whole number: the check still gives x back.

Choose a base and slide xx: the board finds log⁡ax\log_a x by dividing, then raises aa to that power to get xx back. Switch on Swap base and number to see the reciprocal.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q2

Given that log⁡3x−3log⁡x3+2=0\log_3 x - 3\log_x 3 + 2 = 0, find the values of xx.

  1. One base only

    • Swapping the base and the number turns the log over: log⁡x3=1log⁡3x{\log_x 3 = \dfrac{1}{\log_3 x}}.
    • Let y=log⁡3x{y = \log_3 x}. The equation becomes y−3y+2=0{y - \dfrac{3}{y} + 2 = 0}.

    Think first. log₃ x and logₓ 3 have their base and number swapped. How are they related?

  2. Clear the fraction

    • Multiply every term by yy: y2−3+2y=0{y^2 - 3 + 2y = 0}.
    • Put the terms in order: y2+2y−3=0{y^2 + 2y - 3 = 0}.

    Think first. Multiply every term by y.

  3. Solve for y

    • Factorise: (y+3)(y−1)=0{(y + 3)(y - 1) = 0}.
    • So y=1{y = 1} or y=−3{y = -3}.

  4. Back to x

    • log⁡3x=1{\log_3 x = 1} means x=31=3{x = 3^1 = 3}.
    • log⁡3x=−3{\log_3 x = -3} means x=3−3=127{x = 3^{-3} = \frac{1}{27}}.

    Think first. log₃ x = 1 means 3 to what power is x?

Solving log equations

Two moves, usually together:

  1. Combine the logs so there is a single log on each side; then drop the logs: log⁡aP=log⁡aQ\log_a P = \log_a Q means P=QP = Q.
  2. Change to index form when a plain number is left: log⁡aP=n\log_a P = n means P=anP = a^n. A number like 2 can also be written as a log: 2=log⁡101002 = \log_{10} 100.

More: log equations and domains

Two log equations together

Turn each log equation into an ordinary equation, then solve the pair as simultaneous equations. For example, with log⁡x+log⁡y=log⁡12\log x + \log y = \log 12 and log⁡(x−y)=0\log(x - y) = 0:

  • The first equation combines to log⁡xy=log⁡12{\log xy = \log 12}, so xy=12{xy = 12}.
  • The second, in index form, is x−y=100=1{x - y = 10^0 = 1}, so x=y+1{x = y + 1}.
  • Substitute: (y+1)y=12{(y + 1)y = 12}, so y2+y−12=0{y^2 + y - 12 = 0}.
  • Factorise: (y+4)(y−3)=0{(y + 4)(y - 3) = 0}, so y=3{y = 3} or y=−4{y = -4}.
  • Only y=3{y = 3} keeps the logs meaningful. Then x=4{x = 4}.

More: simultaneous log equations

Revision: an exponential equation that hides a quadratic

Your turn

WAEC 2016 · Paper 2 · Q11 (a)

  1. (a)

    Without using mathematical tables or a calculator, solve 2+log⁡10x−log⁡1020=log⁡10(x2+4)2 + \log_{10} x - \log_{10} 20 = \log_{10}(x^2 + 4).

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

  1. Write 2 as a log: 2=log⁡101002 = \log_{10} 100.
  2. The left side is then log⁡10100x20=log⁡105x\log_{10}\dfrac{100x}{20} = \log_{10} 5x.
  3. Drop the logs: 5x=x2+45x = x^2 + 4, so x2−5x+4=0x^2 - 5x + 4 = 0.
  4. Factorise: (x−1)(x−4)=0(x - 1)(x - 4) = 0, so x=1x = 1 or x=4x = 4.
  5. Both make every log positive.

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