Topic: Discriminant question in an IB Math SLE exam disguised as a log question.
Clue: The phrase "exactly one solution" indicates a discriminant problem.
Key Concepts
Discriminant Basics
Discriminant (D): Used to determine the number of solutions in a quadratic equation.
D > 0: Two real solutions.
D = 0: Exactly one real solution (two repeated values).
D < 0: No real solutions (two imaginary solutions).
Problem Statement
Function: ( f(x) = \log_k(x) )
Interval: ( x ) is between 0 and 2.
Given: ( k > 0 ) (base of the logarithm)
Condition: ( f(x) = 2 ) has exactly one solution.
Approach to Problem
Step 1: Recognize the Discriminant Clue
"Exactly one solution" implies setting the discriminant to zero.
Step 2: Manipulate Logarithmic Equation
Convert the log equation to exponential form:
Base ( k ), exponent 2, results in ( k^2 = \text{argument of log} ).
Step 3: Set up a Quadratic Equation
Rearrange to form: ( 3x^2 - 6x + k^2 = 0 ).
Identify coefficients:
( a = 3 )
( b = -6 )
( c = k^2 )
Step 4: Apply Discriminant Formula
Formula: ( b^2 - 4ac = 0 )
Substitute values and solve for ( k ):
( (-6)^2 - 4 \times 3 \times k^2 = 0 )
Simplification: ( 36 = 12k^2 )
( k^2 = 3 )
Step 5: Solve for ( k )
Given ( k > 0 ), solve ( k = \sqrt{3} ).
Conclusion
The solution process involves identifying the discriminant problem, converting log to exponential form, setting up and solving a quadratic equation, and determining the value of ( k = \sqrt{3} ) based on given conditions.