Solve for x log base 5 of 3x+2< log base 5 of 2x+5

Solution:

Step:1

Transform the inequality into an equation: $\log_{5}(3x + 2) = \log_{5}(2x + 5)$.

Step:2

Proceed to solve the resulting equation.

Step:2.1

Equating the logarithmic expressions implies their arguments must be equal: $3x + 2 = 2x + 5$.

Step:2.2

Isolate the variable $x$.

Step:2.2.1

Shift all $x$ terms to one side by subtracting $2x$ from both sides: $3x - 2x + 2 = 5$.

Step:2.2.1.1

Combine like terms: $x + 2 = 5$.

Step:2.2.2

Move constants to the opposite side by subtracting $2$ from both sides: $x = 5 - 2$.

Step:2.2.2.1

Perform the subtraction: $x = 3$.

Step:3

Determine the domain of the function $\log_{5}\left(\frac{3x + 2}{2x + 5}\right)$.

Step:3.1

The argument of the logarithm must be positive: $\frac{3x + 2}{2x + 5} > 0$.

Step:3.2

Solve for the variable $x$.

Step:3.2.1

Identify the zeros of the numerator and denominator: $3x + 2 = 0$ and $2x + 5 = 0$.

Step:3.2.2

Solve the first equation for $x$: $3x = -2$.

Step:3.2.3

Divide by $3$ to isolate $x$: $x = \frac{-2}{3}$.

Step:3.2.4

Solve the second equation for $x$: $2x = -5$.

Step:3.2.5

Divide by $2$ to isolate $x$: $x = \frac{-5}{2}$.

Step:3.2.6

The critical values where the sign of the expression changes are $x = -\frac{2}{3}$ and $x = -\frac{5}{2}$.

Step:3.2.7

Combine these solutions: $x = -\frac{2}{3}, -\frac{5}{2}$.

Step:3.2.8

To find the domain, exclude values that make the denominator zero: $2x + 5 \neq 0$.

Step:3.2.8.1

Solve for $x$: $x \neq -\frac{5}{2}$.

Step:3.2.8.2

The domain is all real numbers except $x = -\frac{5}{2}$.

Step:3.2.9

Create intervals based on the critical values: $x < -\frac{5}{2}$, $-\frac{5}{2} < x < -\frac{2}{3}$, and $x > -\frac{2}{3}$.

Step:3.2.10

Test values from each interval in the original inequality to determine which intervals are part of the solution.

Step:3.2.11

The solution is the union of intervals that satisfy the inequality: $x < -\frac{5}{2}$ or $x > -\frac{2}{3}$.

Step:4

Create test intervals using the critical values and the solution to the equation: $x < -\frac{5}{2}$, $-\frac{5}{2} < x < -\frac{2}{3}$, $-\frac{2}{3} < x < 3$, and $x > 3$.

Step:5

Select test values from each interval and plug them into the original inequality to check which intervals are valid.

Step:6

The solution set includes all intervals that make the inequality true.

Step:7

Express the solution in various forms, such as inequality notation or interval notation.

Knowledge Notes:

1. Logarithmic Inequalities: Inequalities involving logarithms, which can be solved by transforming them into equivalent equations or inequalities without logarithms, considering the domain of the logarithmic function.

2. Domain of a Logarithmic Function: The set of all possible values of $x$ for which the logarithmic function is defined. For $\log_{b}(x)$, the domain is all positive real numbers ($x > 0$).

3. Solving Equations: To find the value(s) of the variable(s) that satisfy an equation.

4. Critical Values: Values of the variable that make the numerator or denominator of a fraction zero. These values are important when determining the sign of a rational expression and its domain.

5. Test Intervals: After finding the critical values, the number line is divided into intervals. Test values from each interval are used to determine the sign of the expression in that interval.

6. Interval Notation: A way to describe sets of numbers by specifying the endpoints of intervals. For example, $(a, b)$ represents all numbers between $a$ and $b$ but not including $a$ or $b$.

7. Inequality Notation: A way to represent the solution to an inequality. For example, $x > a$ means that $x$ is greater than $a$.

8. Undefined Expressions: Expressions that do not have a value in a given context, such as dividing by zero or taking the logarithm of a non-positive number.