Question

1) Find the largest value of x that satisfies:
log5(x2)−log5(x+5)=8

2) Students in a fifth-grade class were given an exam. During the next 2 years, the same students were retested several times. The average score was given by the model
f(t)=90−7log(t+1),    0≤t≤24
where t is the time in months. Round answers to at least 1 decimal point.

A) What is the average score on the original exam?

B) What was the average score after 6 months?

C) What was the average score after 18 months?

Answer 1

1). If log₅(x²)−log₅(x+5)=8 , then log₅ [x²/(x+5)] = 8 ( as log m -log n = log(m/n) ).

Therefore, 5⁸ = x²/(x+5) or, x²/(x+5) = 5⁸ or, x² = 5⁸ (x+5) or, x² -5⁸ x -5⁹ = 0.

Now, on using the quadratic formula, we get x = [5⁸ ±√{ (-5⁸)² -4*1* (-5⁹)}]/2*1 = [5⁸ ±√( 5¹⁶+4*5⁹)]/2 = [5⁸ ± 5³√( 5⁷+4)]/2 = (5³/2)[ 5⁵± √( 5⁷+4)].

Thus, the largest value of x that satisfies the given equation is (5³/2)[ 5⁵+ √( 5⁷+4)] = (125/2([3125+ 3√8681).

2). The average score of the students in a fifth-grade class is given by f(t) = 90−7log(t+1),    0 ≤ t ≤24.

A). When t = 0, we have f(0) = 90−7log(1) = 90. Thus, the average score on the original exam was 90.

B). When t = 6, we have f(6) = 90−7log(7) =90-7*0.8461 = 90-5.92 = 84.1 ( on rounding off to 1 decimal place). Thus, the average score after 6 months was 84.1

C). When t = 18, we have f(18) = 90−7log(19) = 90-7*1.2788= 90-8.95= 81.1 ( on rounding off to 1 decimal place). Thus, the average score after 6 months was 81.1