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}$