Laws of Logarithms (Cambridge International IGCSE Further Maths)

Laws of logarithms

What are the laws of logarithms?

• Laws of logarithms allow you to simplify and manipulate expressions involving logarithms

  • The laws of logarithms are equivalent to the laws of indices

• The laws you need to know are, given $a,x,y>0$:

  • $\mathrm{log}_{a}xy=\mathrm{log}_{a}x+\mathrm{log}_{a}y$

    • This relates to $a^{x}\times a^{y}=a^{x+y}$

  • $\mathrm{log}_{a}\frac{x}{y}=\mathrm{log}_{a}x-\mathrm{log}_{a}y$

    • This relates to $a^{x}\div a^{y}=a^{x-y}$

  • $\mathrm{log}_{a}x^{m}=m\mathrm{log}_{a}x$

    • This relates to $(a^{x})^{y}=a^{xy}$

• There are also some particular results these lead to

  • $\mathrm{log}_{a}a=1$
  • $\mathrm{log}_{a}a^{x}=x$
  • $a^{\mathrm{log}_{a}x}=x$
  • $\mathrm{log}_{a}1=0$
  • $\mathrm{log}_{a}(\frac{1}{x})=-\mathrm{log}_{a}x$

• Beware…

  • …$log_{a} ( x + y ) \text{eq} log_{a} x + log_{a} y$

• These results apply to $\mathrm{ln}x(\mathrm{log}_{e}x)$ too

  • Two particularly useful results are

    • $\mathrm{ln}e^{x}=x$
    • $e^{\mathrm{ln}x}=x$

How do I use the laws of logarithms?

• Laws of logarithms can be used to …

  • … simplify expressions
  • … solve logarithmic equations
  • … solve exponential equations

Exam Hint

• Remember to check whether your solutions are valid

  • log (x+k) is only defined if x > -k
  • You will lose marks if you forget to reject invalid solutions

Worked Example a) Write the expression $2\mathrm{log}4-\mathrm{log}2$ in the form $\mathrm{log}k$, where $k\inℤ$.

Using the law $\mathrm{log}_{a}x^{m}=m\mathrm{log}_{a}x$ we can rewrite $2\mathrm{log}4$ as $\mathrm{log}4^{2}=\mathrm{log}16$.

$2\mathrm{log}4-\mathrm{log}2=\mathrm{log}16-\mathrm{log}2$

Using the law $\mathrm{log}_{a}\frac{x}{y}=\mathrm{log}_{a}x-\mathrm{log}_{a}y$:

• *$\mathrm{log}16-\mathrm{log}2=\mathrm{log}\frac{16}{2}=\mathrm{log}8$

$2\mathrm{log}4-\mathrm{log}2=\mathrm{log}8$

b) Hence, or otherwise, solve $2\mathrm{log}4-\mathrm{log}2=-\mathrm{log}\frac{1}{x}$.

Rewrite the equation using the expression found in part (a).

$2\mathrm{log}4-\mathrm{log}2=\mathrm{log}8$:

$\mathrm{log}8=-\mathrm{log}\frac{1}{x}$

Using the index law $\frac{1}{x}=x^{-1}$::

• *$\mathrm{log}8=-\mathrm{log}x^{-1}$

Using the law $\mathrm{log}_{a}x^{m}=m\mathrm{log}_{a}x$:

$\log 8 = \log x^{- ( - 1 )} = \log x$

Compare the two sides.

• *$\mathrm{log}8=\mathrm{log}x$

$x=8$

Change of Base

How do I change the base of a logarithm?

• The formula for changing the base of a logarithm is

$\mathrm{log}_{a}x=\frac{\mathrm{log}_{b}x}{\mathrm{log}_{b}a}$

• The value you choose for b does not matter, however if you do not have a calculator, you can choose b such that the problem will be possible to solve

Why change the base of a logarithm?

• The laws of logarithms can only be used if the logs have the same base

  • If a problem involves logarithms with different bases, you can change the base of the logarithm and then apply the laws of logarithms

• Changing the base of a logarithm can be particularly useful if you need to evaluate a log problem without a calculator

  • Choose the base such that you would know how to solve the problem from the equivalent exponent

• This formula had more use when calculators were less advanced

  • Some old calculators only had a button for logarithm of base 10

  • To calculate $\mathrm{log}_{5}7$on these calculators you would have to enter

    • $\frac{log_{10} 7}{log_{10} 5}$

• The formula can be useful when evaluating a logarithm where the two numbers are powers of a common number

  • $\mathrm{log}_{4}8=\frac{\mathrm{log}_{2}8}{\mathrm{log}_{2}4}=\frac{3}{2}$

• The formula can be useful when you are solving equations and two logarithms have different bases

  • For example, if you have $\mathrm{log}_{3}k$ and $\mathrm{log}_{9}n$ within the same equation

    • You can rewrite $\mathrm{log}_{9}n$ as $\frac{log_{3} n}{log_{3} 9}$ which simplifies to $\frac{1}{2}\mathrm{log}_{3}n$
    • Or you can rewrite $\mathrm{log}_{3}k$ as $\frac{log_{9} k}{log_{9} 3}$ which simplifies to $2\mathrm{log}_{9}k$

• The formula also allows you to derive and use a formula for switching the numbers:

$log_{a} x = \frac{1}{log_{x} a}$

• Using the fact that $\mathrm{log}_{x}x=1$

Exam Hint

• It is very rare that you will need to use the change of base formula
• Only use it when the bases of the logarithms are different

Worked Example By choosing a suitable value for b, use the change of base law to find the value of $\mathrm{log}_{8}32$ without using a calculator.

Note that 8 and 32 are both powers of 2, where 8 = 2³ and 32 = 2⁵.

*Therefore we can choose b = 2 in the change of base formula. *

$\mathrm{log}_{a}x=\frac{\mathrm{log}_{b}x}{\mathrm{log}_{b}a}$:

$\mathrm{log}_{8}32=\frac{\mathrm{log}_{2}32}{\mathrm{log}_{2}8}$

If 2³= 8 then log₂ 8 = 3 and if 2⁵= 32 then log₂ 32 = 5.

• *$\frac{\mathrm{log}_{2}32}{\mathrm{log}_{2}8}=\frac{5}{3}$

$\mathrm{log}_{8}32=\frac{5}{3}=1\frac{2}{3}$