Explore: Solve Logs with Properties Solutions

Solve Logs with Properties Solutions

Before solving logarithmic equations, it is important to remember the names of each part of the equation and their restrictions.
- Restrictions on the base: $b > 0, b \neq 1$
- Restrictions on the argument: $y > 0$

Check your work to make sure an excluded value is not included as part of the solution because the argument must be greater than zero .

To solve logs of the same base when the equation is equal to…

a single term or number:	one or more logs:
Contract and/or isolate the logs. Write as an exponential equation. Solve. Check.	Contract logs on both sides, as needed. Set arguments equal to one another. Solve. Check.

Example 5

Solve.

$\log_{2} x + \log_{2} (x - 7) = 3$

Plan

Contract logs 

Write as an exponential equation 

Solve

Check

Implement

$\begin{array}{rcl} \log_{2} (x (x - 7)) & = & 3 \\ 2^{3} & = & x (x - 7) \\ 8 & = & x^{2} - 7 x \\ x^{2} - 7 x - 8 & = & 0 \\ (x + 1) (x - 8) & = & 0 \\ x & = & - 1, 8 \end{array}$

Check:

$\log_{2} - 1 + \log_{2} (- 1 - 7) \neq 3$

Note

The argument must be greater than zero.

$\log_{2} 8 + \log_{2} (8 - 7) = 3 ✓$

Example 6

Solve.

$\log_{3} (x + 6) + \log_{3} (x + 4) = \log_{3} 3$

Plan

Contract logs

Set arguments equal to one another

Solve

Check

Implement

Option 1

$\begin{array}{rcl} \log_{3} (x + 6) (x + 4) & = & \log_{3} 3 \\ (x + 6) (x + 4) & = & 3 \\ x^{2} + 10 x + 24 & = & 3 \\ x^{2} + 10 x + 21 & = & 0 \\ (x + 7) (x + 3) & = & 0 \\ x & = & - 7, - 3 \end{array}$

Option 2

$\begin{array}{rcl} \log_{3} (x + 6) (x + 4) & = & 1 \\ (x + 6) (x + 4) & = & 3^{1} \\ x^{2} + 10 x + 24 & = & 3 \end{array}$

Option 3

$\begin{array}{rcl} \log_{3} (x + 6) (x + 4) - \log_{3} 3 & = & 0 \\ \log_{3} \frac{(x + 6) (x + 4)}{3} & = & 0 \\ \frac{x^{2} + 10 x + 24}{3} & = & 3^{0} \\ (3) \frac{x^{2} + 10 x + 24}{3} & = & 1 (3) \\ x^{2} + 10 x + 24 & = & 3 \end{array}$

Check

$\log_{3} (- 7 + 6) + \log_{3} (- 7 + 4) \neq \log_{3} 3$

Note

The argument must be greater than zero.

$\log_{3} (- 3 + 6) + \log_{3} (- 3 + 4) = \log_{3} 3 ✓$

Note

The check can be completed using mental math. If any term results in the argument $< 0$ , then that solution is extraneous.

Example 7

Solve.

$\log_{2} (x - 3) + \log_{2} x = \log_{2} (x + 2) + 2$

$\log_{2} (x - 3) + \log_{2} x - \log_{2} (x + 2) = 2$

$\begin{array}{rcl} \log_{2} \frac{x (x - 3)}{x + 2} & = & 2 \\ 2^{2} & = & \frac{x (x - 3)}{x + 2} \\ 4 (x + 2) & = & x^{2} - 3 x \\ 4 x + 8 & = & x^{2} - 3 x \\ 0 & = & x^{2} - 7 x - 8 \\ (x + 1) (x - 8) & = & 0 \\ x & = & - 1, 8 \end{array}$

Check

$\log_{2} (- 1 - 3) + \log_{2} - 1 - \log_{2} (- 1 + 2) \neq 2$

Note

The argument must be greater than zero.

$\log_{2} (8 - 3) + \log_{2} 8 - \log_{2} (8 + 2) = 2 ✓$

Note

Remember, you can use the cross-product property when working with a proportion.

Example 8

Solve.

$- 4 \log (2 x + 1) + 3 = - 5$

$\begin{array}{rcl} (- \frac{1}{4}) (- 4 \log (2 x + 1)) & = & (- 8) (- \frac{1}{4}) \\ \log (2 x + 1) & = & 2 \\ 10^{2} & = & 2 x + 1 \\ 100 & = & 2 x + 1 \\ 99 & = & 2 x \\ x & = & \frac{99}{2} \end{array}$

Check

$- 4 \log (2 (\frac{99}{2}) + 1) + 3 = - 5 ✓$

Note

The base of a log without a subscript (or base) is 10.