Logarithms Part III

Laws of logarithm

Product Law

$\ \log_a mn = \log_a m + \log_a n $

note that the logarithms are of the same base

Quotient Law

$\ \log_a {\frac{m}{n}} = \log_a m - \log_a n $

note that the logarithms are of the same base

Power Law

$\ \log_a m^n = n\log m $

Important: $\ \log_a xy^n \ne n\log_a xy $

The power n belongs to the base y only. The power n does not belong to base x

$ \begin{aligned}
\log_a xy^n &= \log_a x + \log_a y^n \quad \text{ applying power law} \\
\log_a xy^n &= \log_a x + n\log_a y \\
\end{aligned}$

Using the Laws of logarithms to solve equations

Strategy

1. Obtain a single logarithm function on the LHS as well as the RHS
2. Ensure that logarithm functions on the LHS and RHS have the same base
3. Compare the arguments to solve for x

Example 1

$ \begin{aligned}
\lg {(5y +2)} &= 1 +\lg y \\
\lg {(5y +2)} &= \lg 10 +\lg y \qquad \text{Note that} 1 = \lg10 \\
\lg {(5y +2)} &= \lg 10y \qquad \text{(applying product law)} \\
5y + 2 &= 10y \qquad \text{(comparing arguments)} \\
5y &= 2 \\
y &= \frac{2}{5} \\
\end{aligned}$

Example 2

$ \begin{aligned}
\log_3 {(5y +7)} &= 2 +\log_3 {(y + 4)} \\
\log_3 {(5y +7)} &= 2\log_3{3} + \log_3 {(y + 4)} \qquad \text{Note that} 2 = 2\log_3{3} \\
\lg {(5y +2)} &= \lg 10y \qquad \text{(applying product law)} \\
5y + 2 &= 10y \qquad \text{(comparing arguments)} \\
5y &= 2 \\
y &= \frac{2}{5} \\
\end{aligned}$

Example 3
Solve the following equation

$\log(x+20) - \log(x+2) = \log x $

$\begin{aligned}

\log(x+20) - \log(x+2) &= \log x \\

\log \frac{ x+20}{ x+2 } &= \log x \\

\frac{ x+20}{ x+2 } &= x \\

x + 20 &= x^2 + 2x \\

x^2 + x - 20 &= 0 \\

(x + 5)( x - 4) &= 0 \\

x = -5 \; \text{rejected} \qquad x = 4 \\

\end{aligned} $

$ x = -5 $ is rejected as you cannot log a negative value

Example 4
Solve the following equation: $ 3^{x-4} = 5^{x-1} $

$\begin{aligned}
3^{x-4} &= 5^{x-1} \\
\log{3^{x-4}} &= \log{5^{x-1}} \qquad \text{log both sides} \\
(x-4) \log{3} &= (x-1) \log{5} \\
x( \log{3} - \log{5}) &= 4\log{3} - \log{5} \\
x &= \frac{4\log{3} - \log{5}}{ \log{3} - \log{5}} \\
\end{aligned} $

$ log_{10} 3 $ and $ log_{10} 5 $ can both be evaluated using the calculator


Take home message

To solve log equations:

1. Obtain a single logarithm function on the LHS as well as the RHS
2. Ensure that logarithm functions on the LHS and RHS have the same base
3. Compare the arguments to solve for x

What's Next?
Click here to check out how to solve log equations involving change of base
Click here for Contents Page