Expert: Bobby Soltani - 8/21/2004

Question
I have two questions for you:

1. If 2/3 log 8 + 1/2 log B = 1 - 3 log A. (In this equation in the L.H.S. the base for the log is 5 and in the R.H.S. the base for the is 0.04). Find out the relation between A & B.

2. Consider the sets Tn = {n, n+1, n+2, n+3} where n=1,2,....,97.How many of these sets will not contain 15 or any integral multiple thereof(i.e., any one of the numbers 15,30,45,....)?

Answer
Hi Abir,

1. Remember that loga b = (logc b)/(logc a)
We will use this property where c = 10.

So log5 8 = (log 8)/(log 5)  Since I omitted the base, we assume it is 10.

let's rewrite the equation like this

(2/3)(log 8)/(log 5) + (1/2)(log B)/(log 5) = 1 - 3*(log A)/(log 0.04)
The first term is a constant. It is equal to 0.8614.  The second term we can divide 1/2 by log 5.  This give us,

0.8614 + 0.715*log B = 1 + (2.146)(log A)
subtract 0.8614 from both sides

0.715*log B = 0.1386 + 2.146 *log A

divide both sides by 0.715

log B = 0.194 + 3*log A

take both sides to the power 10

B = 10^(.194 + 3log A)
B = (10^.194)*(10^(3log A)
B = 1.563*(10^(log A^3)
B = 1.563*A^3

That's the answer.

2.  There are a total of 97 sets.  Any set where n = multiple of 15 (m15) or m15-1, or m15-2, or m15-3 will contain m15.  For example, n = 15,14,13, and 12 will all contain 15.  And n = 30,29,28, and 27 will all contain 30.  So we have 4 n's for 15, 30, 45, 60, 75, and 90.  That's 4*6 = 24 n's that will contain m15.  So the number that will not contain m15 is 97-24 = 73.

Let me know if you have questions.  Good luck.

Bobby