### Logarithms Part IV

Changing the base of logarithms

Changing from base a to base c
$\ \log_a b = \frac{\log_c b}{log_c a}$

Changing from base 3 to base 10
$\ \log_3 5 = \frac{\log_{10} 5}{log_{10} 3}$

Solving log equations that involve changing of base

Example 1
Solve the following equation $\log_9 x + \log_3 x = 15$

\begin{aligned} \log_9 x + \log_3 x &= 15 \\ \frac{\log_3 x}{\log_3 9} + \log_3 x &= 15 \qquad \text{changing base} \\ \frac{\log_3 x}{\log_3 3^2} + \log_3 x &= 15 \\ \frac{\log_3 x}{2\log_3 3} + \log_3 x &= 15 \qquad \text{applying power law}\\ \frac{\log_3 x}{2} + \log_3 x &= 10 \qquad \log_3 3 =15 \\ \log_3 x + 2\log_3 x &= 30 \\ \log_3 x + 2\log_3 x &= 30\log_3 3 \qquad 1 = \log_3 3 \\ \log_3 x + \log_3 x^2 &= \log_3 3^{30} \qquad \text{applying power law}\\ \log_3 x^3 &= \log_3 3^{30} \qquad \text{applying product law}\\ x^3 &= 3^{30} \qquad \text{comparing arguments} \\ x &= \sqrt[3] {3^{30}} \\ x &= 3^{10} \\ \end{aligned}

Example 2

Given that $a = \log_{10} 5$ and $b = \log_{10} 3$
Express the following in terms of a and b:
i) $\log_{10} 15$
ii) $\log_{10} 45$
iii) $\log_5 2$

i)
\begin{aligned} \log_{10} 15 &= \log_{10} ( 5 \times 3 ) \\ &= \log_{10} 5 + \log_{10} 3 \\ &= a + b \\ \end{aligned}

iI)
\begin{aligned} \log_{10} 45 &= \log_{10} ( 5 \times 3^2 ) \\ &= \log_{10} 5 + 2\log_{10} 3 \\ &= a + 2b \\ \end{aligned}

iii)
\begin{aligned} \log_5 2 &= \frac{ \log_{10} 2}{ \log_{10} 5} \\ &= \frac{ \log_{10} 2}{ a } \\ &= \frac{ \log_{10} \frac{10}{5} }{ a } \\ &= \frac{ \log_{10} 10 - \log_{10} 5 }{ a } \\ &= \frac{ 1 - a }{ a } \\ \end{aligned}