Algebra 2 Problems of the Day (Algebra 2 Regents, June 2012)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

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Algebra 2/Trigonometry Regents, June 2012

Part IV: A correct answer will receive 6 credits. Partial credit is possible.


39. Solve algebraically for all values of x:

81^{x3 + 2x2} = 27^(5x/3)

Answer:

Both 27 and 81 are powers of 3, so each expression can be rewritten as a power of 3. Once you do that, then the two exponent expressions can be set equal to each other.

81^{x3 + 2x2} = 27^(5x/3)

(3⁴)^{x3 + 2x2} = (3³)^(5x/3)

(3)^{(4)(x3 + 2x2)} = (3)^(3)(5x/3)

(3)^{(4x3 + 8x2)} = (3)^(5x)

4x³ + 8x² = 5x

4x³ + 8x² - 5x = 0

(x)(4x² + 8x - 5) = 0

(x)(4x² + 10x - 2x - 5) = 0

(x) ( (2x)(2x + 5) - 1(2x + 5) ) = 0

(x) (2x - 1)(2x + 5) = 0

x = 0 or 2x - 1 = 0 or 2x + 5 = 0

x = 0 or x = 1/2 or x = -5/2

According to the rubric, writing "(x) (2x - 1)(2x + 5) = 0" was worth only 4 of the 6 points. I would You needed to complete those last two lines to get full credit. One mistake at the end would have gotten you 5 points instead of all 6.

End of Part Exam

More to come. Comments and questions welcome.

More Regents problems.

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