Logarithm equations II

More complex logarithmic equations may require using log rules to combine terms before solving.

Technique: equating arguments

Use log rules to combine multiple log terms into a single log on one or both sides.

After we only have one log on one/both sides,

If $\log_{a} f (x) = \log_{a} g (x),$ then $f (x) = g (x) .$

If $\log_{a} f (x) = c,$ then $f (x) = a^{c} .$

Solve the resulting equation and check for extraneous solutions.

Example: Solving using log rules

Question

Solve the following equations for $x$ :

$\lg x + \lg (x - 3) = 1$
$2 \ln (1 - x) = \ln (2 x + 10) - \ln 2$

Solution

$\begin{matrix} \lg x + \lg (x - 3) & = 1 \\ \lg [x (x - 3)] & = 1 & (Product rule) \\ x (x - 3) & = 10^{1} & (Exponential form) \\ x^{2} - 3 x - 10 & = 0 \\ (x - 5) (x + 2) & = 0 \\ x = 5 & or x = - 2 \end{matrix}$

Since we have $\lg x$ in our question, $x$ must be positive. Thus, we reject $x = - 2.$

The solution is $x = 5. ∎$

$\begin{matrix} 2 \ln (1 - x) & = \ln (2 x + 10) - \ln 2 \\ \ln (1 - x)^{2} & = \ln (\frac{2 x + 10}{2}) & (Power & Quotient rules) \\ \ln (1 - x)^{2} & = \ln (x + 5) \\ (1 - x)^{2} & = x + 5 \\ 1 - 2 x + x^{2} & = x + 5 \\ x^{2} - 3 x - 4 & = 0 \\ (x - 4) (x + 1) & = 0 \\ x = 4 & or x = - 1 \end{matrix}$

Since we have $\ln (1 - x)$ in our question, $1 - x$ must be positive (i.e., $x < 1$ ). Thus, we reject $x = 4.$

The solution is $x = - 1. ∎$