Say I was trying to isolate N and here is my equation:
D <= 3^(2^(N-1))
Is this as simplified as the equation can get?:
((log * (log D / log 3)) + log 2) / log 2 <= N
Here is my logic:
log D <= log 3^(2^(N-1))
log D <= 2^(N-1) * log 3
log D / log 3 <= 2^(N-1)
log * (log D / log 3) <= (N-1)*(log 2)
log * (log D / log 3) <= (N*log2) - log 2
(log * (log D / log 3)) + log 2 <= N*log2
((log * (log D / log 3)) + log 2) / log2 <= N
Did you even read Dr. Peterson's or my posts in your other thread?
There is a family of logarithm functions, not one logarithm function. There are in fact an infinite number of logarithm functions. They share common features but are not identical.
According to pka's view, \(\displaystyle log(a) \equiv log_e(a), \text { where } e \approx 2.718282.\)
because everybody naturally thinks in terms of integral calculus, including kids being introduced to logarithms for the first time.
According to the more old fashioned convention,
\(\displaystyle log(a) \equiv log_{10}(a) \text { and } ln(a) \equiv log_e(a)\)
I find that ancient rule better for a number of reasons such as that I personally find it easier to compute powers of 10. (Hey, I am not a mathematician so I can admit to my reliance on counting by my fingers.)
In either case, for any other base, you
must specify the base of the logarithm. Otherwise, no one has a clue which logarithm function you are talking about. Remember: there are an infinite number of different ones.
\(\displaystyle u = 2^{(n-1)} \text { and } d = 3^u \implies\)
\(\displaystyle log_2(u) = log_2(2^{(n - 1)}) = n - 1 \text { and } \log_3(d) = log_3(3^u) = u \implies\)
\(\displaystyle log_2\{log_3(d)\} = log_2(u) = n - 1 \implies\)
\(\displaystyle n = 1 + log_2\{log_3(d)\}.\)
NOTE: This is a virtual duplicate of my response to the same question asked in another thread.