x2 − 2x − 15 = 0: "2"; Potential solutions are −3 and 5: 5; x2 − 15 = 2x: 1; x − 5 = 0 or x + 3 = 0: 4; (x − 5)(x + 3) = 0: 3

AssignmentFill in the Blanks

Solving Equations using Properties of Logarithms

Question 6 • FL-1200330-Algebra 2 CR

Question text not available

Answer

x2 − 2x − 15 = 0:

"2"

Potential solutions are −3 and 5:

x2 − 15 = 2x:

x − 5 = 0 or x + 3 = 0:

(x − 5)(x + 3) = 0:

Evaluating Logarithmic Expressions

Evaluating Logarithmic Expressions

Evaluating Logarithmic Expressions