Algebra 2 Problems of the Day (Algebra 2/Trigonometry Regents, January 2014)

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

More Regents problems.

Algebra 2/Trigonometry Regents, January 2014

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


39. Solve algebraically for all values of x:

log_{(x + 3)}(2x + 3) + log_{(x + 3)}(x + 5) = 2

Answer:

Have I ever mentioned how much I hated logs as a student. It's likely the notation and trying to memorize a bunch of rules that actually make plenty of sense will you think about the rules for exponents but didn't make a lot of sense for me at the time.

log_{(x + 3)}(2x + 3) + log_{(x + 3)}(x + 5) = 2
log_{(x + 3)}((2x + 3)(x + 5)) = 2
log_{(x + 3)}(2x² + 13x + 15) = 2
(x + 3)² = 2x² + 13x + 15
x² + 6x + 9 = 2x² + 13x + 15
x² + 7x + 6 = 0
(x + 6)(x + 1) = 0
x + 6 = 0 or x + 1 = 0
x = -6 or x = -1

We reject x = -6 because the base x + 3 would be (-6) + 3 = -3, which is not allowed.

So x = -1 only.

End of exam.

More to come. Comments and questions welcome.

More Regents problems.

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