Comments on: 1536= 3(2)^n-1 using log? http://www.about-siding.com/1536-32n-1-using-log/130/ Your Questions, Our Answers Mon, 21 May 2012 04:16:13 +0000 http://wordpress.org/?v=2.2.1 By: michaelempeigne http://www.about-siding.com/1536-32n-1-using-log/130/#comment-213 michaelempeigne Wed, 30 Apr 2008 15:14:00 +0000 http://www.about-siding.com/1536-32n-1-using-log/130/#comment-213 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 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

]]> By: ale_23 http://www.about-siding.com/1536-32n-1-using-log/130/#comment-212 ale_23 Sun, 27 Apr 2008 21:32:49 +0000 http://www.about-siding.com/1536-32n-1-using-log/130/#comment-212 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 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

]]> By: Crouching Doggie http://www.about-siding.com/1536-32n-1-using-log/130/#comment-211 Crouching Doggie Sun, 27 Apr 2008 04:33:54 +0000 http://www.about-siding.com/1536-32n-1-using-log/130/#comment-211 Simply use antilog to find the answer. Simply use antilog to find the answer.

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