Home > Just Math
| Real Powerful Logarithms (Posted on 2008-04-04) |
 |
Determine all possible positive real pair(s) (A, B) that satisfy the following system of simultaneous equations:
(3A)log 3 = (8B)log 2, and:
2log A = 3log B * 4log 3
Note: For the purposes of the problem, the base of the logarithm is a positive real number > 1.
|
|
Submitted by K Sengupta
|
|
Rating: 5.0000 (1 votes)
|
|
|
Solution:
|
(Hide)
|
(A, B) = (1/3, 1/8) is the only possible solution.
EXPLANATION:
Taking logs on both sides of the second equation, we have:
(log A – 2*log 3)*log 2 = logB* log 3
or, (log A/log 3) – 2 = log B/log 2 = m (say)
or, log A = (m+2)*log 3 and, log B = m*log 2
Taking antilogs on both sides, we have:
A = 3m+2, and B = 2m
Substituting these values in the first equation, we have:
3((m+3)*log 3) = 2((m+3)*log 2)
or, (m+3)*((log 3)2 – (log 2)2)) = 0 ...(*)
Since the base of the logarithm is > 1, we must have:
(log3)2 > (log 2)2
Thus, from (*), we must have:
m+3 = 0, giving m=-3.
Consequently, A = 3-1 = 1/3, and B = 2-3 = 1/8
------------------------------------
For an alternative methodology, refer to the solution submitted by Paul in this location.
|
Comments: (
You must be logged in to post comments.)
|
|
Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|
