1536= 3(2)^n-1 using log?
Posted by admin
Miguel O asked:
so the question is, how do i use log to solve 1536=3 X 2^n-1
so the question is, how do i use log to solve 1536=3 X 2^n-1
i found out nn = 10, but i want to know how log can be used to solve this equation.
i know you log both sides, but how is it gonna look like, and how iis it solved
James
April 27th, 2008 at 12:33 am
Simply use antilog to find the answer.
April 27th, 2008 at 5:32 pm
1536 = 3(2)^(n - 1)
1536 / 3 = 2^(n - 1)
512 = 2^(n - 1)
log_2 (512) = log_2 [2^(n - 1)]
log_2 (2^9) = log_2 [2^(n - 1)]
9 log_2 (2) = (n - 1) log_2 (2)
9 = n - 1
n = 10
Alejandra
April 30th, 2008 at 11:14 am
log 1536 = log 3 + log (2)^(n-1)
log 1536 - log 3 = (n-1) log 2
log (1536/3) = n log 2 - log 2
log (1536/3) + log 2 = n log 2
log (512) + log 2 = n log 2
log 1024 / log 2 = n
10 = n