Topic 7 Logarithms

If \(a=b^x\), then \(x\) is called the logarithm of \(a\) to the base \(b\). In symbols \(x=\log_b a\).

For example, \(\log_3 9=2\),     \(\log_2 32=5\),     \(\log_{10} 1000=3\).

We require the base \(b\) to be a positive number, not equal to \(1\).

Notation for two most commonly used bases:

  • \(\log(a) := \log_{10}(a)\)     (used in Engineering)
  • \(\ln(a) := \log_e(a)\)     where \(e\) is the Euler’s number, \(e \approx 2.7183\) (to 4 d.p.)
  • \(\ln(a)\) is also called the natural logarithm of \(a\).

Note:: \(\log_{10}(a) \begin{cases} =0 & \text{for }a=1\\ >0& \text{for }a>1\\ <0 &\text{for }0<a<1\\\text{does not exist}&\text{for }a\leq0.\end{cases}\)

Graph of the natural logarithm function

Figure 7.1: Graph of the natural logarithm function

7.1 Laws of logarithms

  1. \(\log_b(M\cdot N)=\log_b(M)+\log_b(N)\)

  2. \(\log_b(N^a)=a\cdot\log_b(N)\)

Proof (non-examinable):

  1. \(\log_b(M\cdot N)\) is defined by the equation \(b^{\log_b(M\cdot N)}=M\cdot N\).
    So, we only need to verify that \(b^{\log_b(M)+\log_b(N)}=M\cdot N\).
    But by laws of exponents and the definition of logarithm, we have LHS \(= b^{\log_b(M)}\cdot b^{\log_b(N)}=M\cdot N\), so we are done.

  2. \(\log_b(N^a)\) is defined by the equation \(b^{\log_b(N^a)}=N^a\).
    So, we only need to verify that \(b^{a\cdot \log_b(N)}=N^a\).
    But by laws of exponents and the definition of logarithm, we have LHS \(= \left(b^{\log_b(N)}\right)^a= N^a\), so we are done.

From 1. and 2. we can derive two more laws:

  1. \(\log_b\left(\frac{M}{N}\right) = \log_b(M)-\log_b(N)\)

  2. \(\log_b(N) = \dfrac{\log_{10}(N)}{\log_{10}(b)}\) \(=\dfrac{\ln(N)}{\ln(b)} = \dfrac{\log_c(N)}{\log_c(b)}\) for any base \(c\).

Proof (non-examinable):

  1. LHS \(= \log_b(M\cdot N^{-1}) \stackrel{1.}{=} \log_b(M) + \log_b(N^{-1}) \stackrel{2.}{=} \log_b(M)-\log_b(N) =\) RHS

  2. We apply \(\log_c\) to both sides of the equation \(b^{\log_b(N)}=N\): \[\begin{align*} \log_c\left(b^{\log_b(N)}\right) &= \log_c(N) & \\ \log_b(N)\cdot \log_c(b) &= \log_c(N) & \text{(by 2.)}\\ \log_b(N) &= \frac{\log_c(N)}{\log_c(b)} & (\text{dividing by }\log_c(b)\not=0) \end{align*}\]

Example 7.1 \(\log_5(2.623) \stackrel{4.}{=} \dfrac{\log_{10}(2.623)}{\log_{10}(5)} \approx \dfrac{0.4188}{0.6990} \approx 0.5991\)

Example 7.2 Solve \(2^x =5\).

Solution. \(x=\log_2(5)=\dfrac{\log_{10}(5)}{\log_{10}(2)} \approx\dfrac{0.6990}{0.3010}\approx 2.322\)

Example 7.3 Solve \(3^{x+1}=2^{2x-3}\).

Solution. Applying \(\log_{10}\) on both sides: \(\log_{10}(3^{x+1}) = \log_{10}(2^{2x-3})\).
\(\implies\) \((x+1)\log_{10}(3) = (2x-3)\log_{10}(2)\)
\(\implies\) \(x\left(\log_{10}(3)-2\log_{10}(2)\right) = -3\log_{10}(2)-\log_{10}(3)\)
\(\implies\) \(x=\dfrac{-3\log_{10}(2)-\log_{10}(3)}{\log_{10}(3)-2\log_{10}(2)}\approx 11.0471\)

Example 7.4 Solve \(2^{2x}-2^{x+1}-15=0\).

Solution. \(2^{2x}-2^{x+1}-15=0\)
\(\iff\) \(\left(2^x\right)^2-2\cdot\left(2^x\right)-15=0\)  (by laws of exponents)
\(\iff\) \(y^2 -2y-15=0\)  (where \(y=2^x\))
\(\iff\) \((y-5)(y+3)=0\)  (factorising)
\(\iff\) \(y=5\) or \(y=-3\)
\(\iff\) \(x=\log_2(5)\approx 2.322\),
and \(2^x=-3\) is not possible (so no solution)